# questions

tags: ['Physics', 'Electromagnetic Induction', 'BITSAT', 'BITSAT 2023 (Memory Based Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-conducting-loop-of-radius-10-cm-is-placed-perpendicular-to-a-uniform-magnetic-field-of-05-t-the-magnetic-field-is-decreased-to-zero-in-05-s-at-a-ste

slug:
a-conducting-loop-of-radius-10-cm-is-placed-perpendicular-to-a-uniform-magnetic-field-of-05-t-the-magnetic-field-is-decreased-to-zero-in-05-s-at-a-ste-84606

question:
<div class="question">A conducting loop of radius $\frac{10}{\sqrt{\pi}} \mathrm{cm}$ is placed perpendicular to a uniform magnetic field of $0.5 T$. The magnetic field is decreased to zero in $0.5 s$ at a steady rate. The induced emf in the circular loop at $0.25 s$ is:</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{emf}=1 \mathrm{mV}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\mathrm{emf}=10 \mathrm{mV}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{emf}=100 \mathrm{mV}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{emf}=5 \mathrm{mV}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{emf}=10 \mathrm{mV}$</span> </div>

answer:
<div class="solution">As $\left.\varepsilon\right|_{t=0.5 \mathrm{sec}}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$<br/>$=-\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \quad\left[\because \theta=0^{\circ} \Rightarrow \cos \theta=1\right]$<br/>$=-\pi \times\left(\frac{10}{\sqrt{\pi}}\right)^2 \times 10^{-4} \times \frac{0-0.5}{0.5}=10^{-2} \mathrm{~V}=10 \mathrm{mV}$<br/>As $\frac{d B}{d t}=$ constant $\Rightarrow$ Induced emf will not change with time. So, $\left.\mathrm{e}\right|_{0,5 \mathrm{sec}}=\left.\mathrm{e}\right|_{0.25 \mathrm{sec}}=10 \mathrm{mV}$</div>


tags: ['Physics', 'Electromagnetic Induction', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-conducting-square-loop-is-placed-in-a-magnetic-field-b-with-its-plane-perpendicular-to-the-field-the-sides-of-the-loop-start-shrinking-at-a-constant

slug:
a-conducting-square-loop-is-placed-in-a-magnetic-field-b-with-its-plane-perpendicular-to-the-field-the-sides-of-the-loop-start-shrinking-at-a-constant-92571

question:
<div class="question">A conducting square loop is placed in a magnetic field $B$ with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate $\alpha$. The induced emf in the loop at an instant when its side is ' $a$ ' is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$2 a \alpha B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$a^{2} \alpha B$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$2 a^{2} \alpha B$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$a \alpha B$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$2 a \alpha B$</span> </div>

answer:
<div class="solution">At any time $t$, the side of the square $a=\left(a_{0}\right.$ $-\alpha t),$ where $a_{0}=$ side at $t=0$
<br/> <br/>At this instant, flux through the square :
<br/> <br/>$$
<br/> <br/>\begin{array}{l}
<br/> <br/>\phi=B A \cos 0^{\circ}=B\left(a_{0}-\alpha t\right)^{2} \\
<br/> <br/>\therefore \text { emf induced } E=-\frac{d \phi}{d t} \\
<br/> <br/>\Rightarrow E=-B .2\left(a_{0}-\alpha t\right)(0-\alpha)=+2 \alpha a B
<br/> <br/>\end{array}
<br/> <br/>$$</div>


tags: ['Physics', 'Electromagnetic Induction', 'BITSAT', 'BITSAT 2020']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">zero</span> </div>
source_db: MarksBatch1_P1.db
title:
a-conducting-square-loop-of-side-l-and-resistance-r-moves-in-its-plane-with-a-uniform-velocity-v-perpendicular-to-one-of-its-side-a-magnetic-induction

slug:
a-conducting-square-loop-of-side-l-and-resistance-r-moves-in-its-plane-with-a-uniform-velocity-v-perpendicular-to-one-of-its-side-a-magnetic-induction-77363

question:
<div class="question">A conducting square loop of side $\mathrm{L}$ and resistance $\mathrm{R}$ moves in its plane with a uniform velocity v perpendicular to one of its side. A magnetic induction $\mathrm{B}$ constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/v3-1Mt3mK4D2OS5PERmiMAk_YvGxnqyAuKIeBkon8vw.original.fullsize.png"/><br/> <br/> <br/>The current induced in the loop is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$ clockwise</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}$ anticlockwise</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{2 \mathrm{~B} \ell v}{\mathrm{R}}$ anticlockwise</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">zero</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

answer:
<div class="solution">Since the magnetic field is uniform the flux $f$ through the square loop at any time $t$ is constant, because
<br/> <br/>$\begin{array}{l}
<br/> <br/>\mathrm{f}=\mathrm{B} \times \mathrm{A}=\mathrm{B} \times \mathrm{L}^{2}=\text { constant } \\
<br/> <br/>\therefore \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\text { zero }
<br/> <br/>\end{array}$</div>


tags: ['Physics', 'Rotational Motion', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-constant-torque-of-314-n-m-is-exerted-on-a-pivoted-wheel-if-angular-acceleration-of-wheel-is-4-rad-s-2-then-the-moment-of-inertia-of-the-wheel-is

slug:
a-constant-torque-of-314-n-m-is-exerted-on-a-pivoted-wheel-if-angular-acceleration-of-wheel-is-4-rad-s-2-then-the-moment-of-inertia-of-the-wheel-is-27177

question:
<div class="question">A constant torque of $31.4 \mathrm{~N}-\mathrm{m}$ is exerted on a pivoted wheel. If angular acceleration of wheel is $4 \pi \mathrm{rad} / \mathrm{s}^{2},$ then the moment of inertia of the wheel is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$2.5 \mathrm{~kg} \mathrm{~m}^{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$3.5 \mathrm{~kg} \mathrm{~m}^{2}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$4.5 \mathrm{~kg} \mathrm{~m}^{2}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$5.5 \mathrm{~kg} \mathrm{~m}^{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$2.5 \mathrm{~kg} \mathrm{~m}^{2}$</span> </div>

answer:
<div class="solution">$I=\frac{\tau}{\alpha}=\frac{31.4}{4 \pi}=\frac{31.4}{4 \times 3.14}=2.5 \mathrm{~kg} \mathrm{~m}^{2}$.</div>


tags: ['Physics', 'Current Electricity', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-constant-voltage-is-applied-between-the-two-ends-of-a-uniform-metallic-wire-some-heat-is-developed-in-it-the-heat-developed-is-doubled-if

slug:
a-constant-voltage-is-applied-between-the-two-ends-of-a-uniform-metallic-wire-some-heat-is-developed-in-it-the-heat-developed-is-doubled-if-72029

question:
<div class="question">A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">both the length and the radius of the wire are halved.</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">both the length and the radius of the wire are doubled.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">the radius of the wire is doubled.</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">the length of the wire is doubled.</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">both the length and the radius of the wire are doubled.</span> </div>

answer:
<div class="solution">The heat produced is given by
<br/> <br/>$$
<br/> <br/>H=\frac{V^{2}}{R} \text { and } R=\frac{\rho \ell}{\pi r^{2}}
<br/> <br/>$$
<br/> <br/>$\therefore H=V^{2}\left(\frac{\pi r^{2}}{\rho \ell}\right)$
<br/> <br/>or $\quad H=\left(\frac{\pi V^{2}}{\rho}\right) \frac{r^{2}}{\ell}$
<br/> <br/>Thus heat $(H)$ is doubled if both length $(\ell)$ and radius $(r)$ are doubled.</div>


tags: ['Physics', 'Ray Optics', 'BITSAT', 'BITSAT 2019']
source_db: MarksBatch1_P1.db
title:
a-convex-lens-of-focal-length-25-cm-produces-images-of-the-same-magnification-2-when-an-object-is-kept-at-two-positions-x-1-and-x-2-x-1-x-2-from-the-l

slug:
a-convex-lens-of-focal-length-25-cm-produces-images-of-the-same-magnification-2-when-an-object-is-kept-at-two-positions-x-1-and-x-2-x-1-x-2-from-the-l-67361

question:
<div class="question">A convex lens of focal length <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>25</mn><mo> </mo><mi>cm</mi></math> produces images of the same magnification <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn></math>, when an object is kept at two positions <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mn>1</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mn>2</mn></msub></math> <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><msub><mi>x</mi><mn>1</mn></msub><mo>&gt;</mo><msub><mi>x</mi><mn>2</mn></msub></mrow></mfenced></math> from the lens. The ratio of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mn>2</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mn>1</mn></msub></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo>:</mo><mn>1</mn></math></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>:</mo><mn>3</mn></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>:</mo><mn>2</mn></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>:</mo><mn>1</mn></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>:</mo><mn>3</mn></math></span> </div>

answer:
<div class="solution"><p>For the given convex lens, focal length, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mn>25</mn><mo> </mo><mi>cm</mi></math></p><p>Magnification of lens in terms of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>u</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi></math> is given by,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mo>=</mo><mfrac><mi>f</mi><mrow><mi>f</mi><mo>+</mo><mi>u</mi></mrow></mfrac><mo>=</mo><mfrac><mn>25</mn><mrow><mn>25</mn><mo>+</mo><mi>u</mi></mrow></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mi mathvariant="normal">i</mi></mfenced></math></p><p>When the image is real, then <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mo>=</mo><mo>-</mo><mn>2</mn></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>u</mi><mo>=</mo><mo>-</mo><msub><mi>x</mi><mn>1</mn></msub></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo></math> From Eq. <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mi mathvariant="normal">i</mi><mo>)</mo></math>, <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>2</mn><mo>=</mo><mfrac><mn>25</mn><mrow><mn>25</mn><mo>-</mo><msub><mi>x</mi><mn>1</mn></msub></mrow></mfrac><mo>⇒</mo><msub><mi>x</mi><mn>1</mn></msub><mo>=</mo><mfrac><mn>75</mn><mn>2</mn></mfrac></math></p><p>Again, when the image is virtual, then <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mo>=</mo><mn>2</mn></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>u</mi><mo>=</mo><mo>-</mo><msub><mi>x</mi><mn>2</mn></msub></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo></math> From Eq. <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mi mathvariant="normal">i</mi><mo>)</mo></math>, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo>=</mo><mfrac><mn>25</mn><mrow><mn>25</mn><mo>-</mo><msub><mi>x</mi><mn>2</mn></msub></mrow></mfrac><mo>⇒</mo><msub><mi>x</mi><mn>2</mn></msub><mo>=</mo><mfrac><mn>25</mn><mn>2</mn></mfrac></math></p><p>Hence, the ratio of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mn>2</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mn>1</mn></msub></math> is</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mfrac><msub><mi>x</mi><mn>2</mn></msub><msub><mi>x</mi><mn>1</mn></msub></mfrac><mo>=</mo><mfrac><mrow><mn>25</mn><mo>/</mo><mn>2</mn></mrow><mrow><mn>75</mn><mo>/</mo><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac></math></p></div>


tags: ['Physics', 'Ray Optics', 'BITSAT', 'BITSAT 2018']
source_db: MarksBatch1_P1.db
title:
a-convex-lens-shown-in-the-figure-is-made-up-of-two-types-of-transparent-materials-a-point-sources-of-light-is-placed-on-its-principal-axis-if-reflect

slug:
a-convex-lens-shown-in-the-figure-is-made-up-of-two-types-of-transparent-materials-a-point-sources-of-light-is-placed-on-its-principal-axis-if-reflect-91699

question:
<div class="question">A convex lens shown in the figure is made up of two types of transparent materials. A point sources of light is placed on its principal axis. If reflections from the boundaries between layers are ignored, the lens will form<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/9d490464-232a-4a92-a368-eb213a0feddd-image.png" style="width: 90px; height: 261px;"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">only one image</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">two images</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">infinite images</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">no image at all</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">two images</span> </div>

answer:
<div class="solution">Since, the lens is made up of two kinds of transparent material, it has two refractive indices for the incident beam of light. Hence, there will be two focal lengths of the lens and therefore two images will be observed.</div>


tags: ['Physics', 'Magnetic Effects of Current', 'BITSAT', 'BITSAT 2018']
source_db: MarksBatch1_P1.db
title:
a-copper-rod-of-mass-m-slides-under-gravity-on-two-smooth-parallel-rails-l-distance-apart-and-set-an-angle-to-the-horizontal-at-the-bottom-the-rails-a

slug:
a-copper-rod-of-mass-m-slides-under-gravity-on-two-smooth-parallel-rails-l-distance-apart-and-set-an-angle-to-the-horizontal-at-the-bottom-the-rails-a-59204

question:
<div class="question"><p>A copper rod of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> slides under gravity on two smooth parallel rails <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi></math> distance apart and set an angle <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>θ</mi></math> to the horizontal. At the bottom, the rails are joined by a resistance <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi></math> in figure. There is a uniform magnetic field <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi></math> perpendicular to the plane of the rails. The terminal velocity of rod is</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/f9f90ec0-b008-4e91-bc69-94f4537499cc-image.png" style="width: 200px; height: 181px;"/></p></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi><mi>R</mi><mi>tan</mi><mi>θ</mi></mrow><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>l</mi><mn>2</mn></msup></mrow></mfrac></math></p></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi><mi>R</mi><mi>cot</mi><mi>θ</mi></mrow><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>l</mi><mn>2</mn></msup></mrow></mfrac></math></p></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi><mi>R</mi><mi>sin</mi><mi>θ</mi></mrow><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>l</mi><mn>2</mn></msup></mrow></mfrac></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi><mi>R</mi><mi>cos</mi><mi>θ</mi></mrow><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>l</mi><mn>2</mn></msup></mrow></mfrac></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi><mi>R</mi><mi>sin</mi><mi>θ</mi></mrow><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>l</mi><mn>2</mn></msup></mrow></mfrac></math></span> </div>

answer:
<div class="solution"><p>Terminal velocity of the rod is attained when magnetic force on the rod (Bil) balances the component of weight of the rod<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mi>m</mi><mi>g</mi><mi>sin</mi><mi>θ</mi><mo>)</mo><mo>,</mo></math> as in figure.</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/fb0451a0-a03a-4b9e-b023-766ce25f29ea-image.png" style="width: 240px; height: 180px;"/></p><p>So, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Bil</mi><mo>=</mo><mi>m</mi><mi>g</mi><mi>sin</mi><mi>θ</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi><mfenced><mfrac><mi>e</mi><mi>R</mi></mfrac></mfenced><mi>l</mi><mo>=</mo><mi>m</mi><mi>g</mi><mi>sin</mi><mi>θ</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>B</mi><mi>l</mi><mi>e</mi></mrow><mi>R</mi></mfrac><mo>=</mo><mi>m</mi><mi>g</mi><mi>sin</mi><mi>θ</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>B</mi><mi>l</mi><mfenced><mrow><mi>B</mi><mo>/</mo><msub><mi>V</mi><mi>T</mi></msub></mrow></mfenced></mrow><mi>R</mi></mfrac><mo>=</mo><mi>m</mi><mi>g</mi><mi>sin</mi><mi>θ</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>v</mi><mi>T</mi></msub><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi><mi>sin</mi><mi>θ</mi><mi>R</mi></mrow><mrow><msup><mi>B</mi><mn>2</mn></msup><msup><mi>l</mi><mn>2</mn></msup></mrow></mfrac></math></p></div>


tags: ['Physics', 'Thermal Properties of Matter', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-copper-sphere-cools-from-6-2-c-to-5-0-c-in-10-minutes-and-to-4-2-c-in-the-next-10-minutes-calculate-the-temperature-of-the-surroundings

slug:
a-copper-sphere-cools-from-6-2-c-to-5-0-c-in-10-minutes-and-to-4-2-c-in-the-next-10-minutes-calculate-the-temperature-of-the-surroundings-94839

question:
<div class="question">A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surroundings.</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$18.01^{\circ} \mathrm{C}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$26^{\circ} \mathrm{C}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$10.6^{\circ} \mathrm{C}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$20^{\circ} \mathrm{C}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$26^{\circ} \mathrm{C}$</span> </div>

answer:
<div class="solution">By Newton's law of cooling, $\frac{\theta_{1}-\theta_{2}}{t}=-k\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ ....(1)<br/><br/>A sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. $\frac{62-50}{10}=-\mathrm{k}\left[\frac{62+50}{2}-\theta_{0}\right]$ ....(2)<br/>Now, sphere cools from $50^{\circ} \mathrm{C}$ to $42^{\circ} \mathrm{C}$ in next $10 \mathrm{~min} .$<br/>$$<br/>\frac{50-42}{10}=-\mathrm{k}\left[\frac{50+42}{2}-\theta_{0}\right] ....(3)<br/>$$<br/>Dividing eq $^{\text {n }}$. (2) by (3) we get, $\frac{56-\theta_{0}}{46-\theta_{0}}=\frac{1.2}{0.8}$ or $0.4 \theta_{0}=10.4$<br/>Hence $\theta_{0}=26^{\circ} \mathrm{C}$</div>


tags: ['Physics', 'Thermal Properties of Matter', 'BITSAT', 'BITSAT 2020']
source_db: MarksBatch1_P1.db
title:
a-crystal-has-a-coefficient-of-expansion-13-1-0-7-in-one-direction-and-231-1-0-7-in-every-direction-at-right-angles-to-it-then-the-cubical-coefficient

slug:
a-crystal-has-a-coefficient-of-expansion-13-1-0-7-in-one-direction-and-231-1-0-7-in-every-direction-at-right-angles-to-it-then-the-cubical-coefficient-34539

question:
<div class="question">A crystal has a coefficient of expansion $13 \times 10^{-7}$ in one direction and $231 \times 10^{-7}$ in every direction at right angles to it. Then the cubical coefficient of expansion is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$462 \times 10^{-7}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$244 \times 10^{-7}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$475 \times 10^{-7}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$257 \times 10^{-7}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$475 \times 10^{-7}$</span> </div>

answer:
<div class="solution">$\begin{aligned} \gamma &amp;=\alpha_{1}+\alpha_{2}+\alpha_{3} \\ &amp;=13 \times 10^{-7}+231 \times 10^{-7}+231 \times 10^{-7} \\ &amp;=475 \times 10^{-7} \end{aligned}$</div>


tags: ['Physics', 'Mechanical Properties of Solids', 'BITSAT', 'BITSAT 2013']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">0.06</span> </div>
source_db: MarksBatch1_P1.db
title:
a-cube-is-subjected-to-a-uniform-volume-compression-if-the-side-of-the-cube-decreases-by-2-the-bulk-strain-is

slug:
a-cube-is-subjected-to-a-uniform-volume-compression-if-the-side-of-the-cube-decreases-by-2-the-bulk-strain-is-88866

question:
<div class="question">A cube is subjected to a uniform volume compression. If the side of the cube decreases by $2 \%$ the bulk strain is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">0.02</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">0.03</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">0.04</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">0.06</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

answer:
<div class="solution">\(\begin{aligned} \frac{\Delta a}{d} &amp;=2 \% \\ v &amp;=a^{3} \\ \frac{\Delta v}{v} &amp;=3 \frac{\Delta a}{d} \\ &amp;=6 \% \\ &amp;=0.06 \end{aligned}\)</div>


tags: ['Chemistry', 'Electrochemistry', 'BITSAT', 'BITSAT 2013']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">3</span> </div>
source_db: MarksBatch1_P1.db
title:
a-current-of-20-a-passed-for-5-hours-through-a-molten-metal-salt-deposits-222-g-of-metal-at-wt-177-the-oxidation-state-of-the-metal-in-the-metal-salt-

slug:
a-current-of-20-a-passed-for-5-hours-through-a-molten-metal-salt-deposits-222-g-of-metal-at-wt-177-the-oxidation-state-of-the-metal-in-the-metal-salt-10031

question:
<div class="question">A current of 2.0 A passed for 5 hours through a molten metal salt deposits $22.2 \mathrm{~g}$ of metal (At wt. $=177$ ). The oxidation state of the metal in the metal salt is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">1</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">2</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">3</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">4</span> </li> </ul>

answer:
<div class="solution">$\mathrm{m}=\frac{\mathrm{E} \cdot \mathrm{wt} \times \mathrm{Q}}{96500}$
<br/> <br/>$\begin{aligned} \therefore \mathrm{E} . \mathrm{wt} &amp;=\frac{\mathrm{m} \times 96500}{\mathrm{Q}} \\=\frac{22.2 \times 96500}{2 \times 5 \times 60 \times 60}=60.3 \end{aligned}$
<br/> <br/>Oxidation state $=\frac{\text { At wt. }}{\text { Eq. wt. }}=\frac{177}{60.3}=3$</div>


tags: ['Physics', 'Thermodynamics', 'BITSAT', 'BITSAT 2009']
answer: <div class="solution">No Solution available</div>
source_db: MarksBatch1_P1.db
title:
a-cyclic-process-is-shown-in-the-pt-diagram-which-of-the-curves-show-the-same-process-on-a-p-v-diagram

slug:
a-cyclic-process-is-shown-in-the-pt-diagram-which-of-the-curves-show-the-same-process-on-a-p-v-diagram-65283

question:
<div class="question">A cyclic process is shown in the p-T diagram. Which of the curves show the same process on a $\mathrm{P}$ - $\mathrm{V}$ diagram?<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/o-8UK9Po54w8g9dEv-1EPuWhAq8UOM4iruEEa46yXHY.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/4Yvdldf6HPAimaHiWd2_XC-ufb25mh5CJgIjXJNi6ik.original.fullsize.png"/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/1dLNjBpUEZ-s3gQOzMfxU_mUCnVbc5IKgr3Wz65oSrk.original.fullsize.png"/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/quZEfbvg6niTaruX23KL8HkrNiMHDASqDRS9pa8ZE-4.original.fullsize.png"/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/mh8iXgW4bMwNmhykyKagHV5Ts_TyOvJRul4MbGSMjuA.original.fullsize.png"/></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/1dLNjBpUEZ-s3gQOzMfxU_mUCnVbc5IKgr3Wz65oSrk.original.fullsize.png"/></span> </div>


tags: ['Physics', 'Rotational Motion', 'BITSAT', 'BITSAT 2019']
source_db: MarksBatch1_P1.db
title:
a-cylinder-of-mass-2-kg-is-released-from-rest-from-the-top-of-an-inclined-plane-of-inclination-30-and-length-1-m-if-the-cylinder-rolls-without-slippin

slug:
a-cylinder-of-mass-2-kg-is-released-from-rest-from-the-top-of-an-inclined-plane-of-inclination-30-and-length-1-m-if-the-cylinder-rolls-without-slippin-73086

question:
<div class="question">A cylinder of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo> </mo><mi>kg</mi></math> is released from rest from the top of an inclined plane of inclination <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>30</mn><mo>°</mo></math> and length <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi mathvariant="normal">m</mi></math>. If the cylinder rolls without slipping, then its speed when it reaches the bottom is [<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>g</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>2</mn></mrow></msup></math>]</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>20</mn><mn>3</mn></mfrac><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mfrac><mn>20</mn><mn>3</mn></mfrac></msqrt><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>10</mn><mn>3</mn></mfrac><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mfrac><mn>10</mn><mn>3</mn></mfrac></msqrt><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mfrac><mn>20</mn><mn>3</mn></mfrac></msqrt><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </div>

answer:
<div class="solution"><p>Potential energy of cylinder at top position</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/610ef1cd-4773-43f2-8366-3b0bd8d1f9fa-image.png" style="width: 230px; height: 177px;"/></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>U</mi><mo>=</mo><mi>m</mi><mi>g</mi><mfenced><mrow><mi>A</mi><mi>B</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>2</mn><mo>×</mo><mn>10</mn><mfenced close="]" open="["><mrow><mn>1</mn><mo>.</mo><mi>sin</mi><mn>30</mn><mo>°</mo></mrow></mfenced><mo>  </mo><mfenced close="]" open="["><mtable columnalign="left"><mtr><mtd><mo>∵</mo><mi>sin</mi><mn>30</mn><mo>°</mo><mo>=</mo><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>B</mi><mi>C</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>∴</mo><mi>A</mi><mi>B</mi><mo>=</mo><mn>1</mn><mi>sin</mi><mn>30</mn><mo>°</mo></mtd></mtr></mtable></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">J</mi></math></p><p>If <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi></math> is the linear speed of cylinder when it reaches at bottom, then at bottom, total potential energy of cylinder is converted into kinetic energy,</p><p>i.e. <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi><mo>=</mo><mi>U</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>I</mi><msup><mi>ω</mi><mn>2</mn></msup><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>m</mi><msup><mi>v</mi><mn>2</mn></msup><mo>=</mo><mn>10</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>·</mo><mfrac><mrow><mi>m</mi><msup><mi>r</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac><mo>·</mo><mfrac><msup><mi>v</mi><mn>2</mn></msup><msup><mi>r</mi><mn>2</mn></msup></mfrac><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>m</mi><msup><mi>v</mi><mn>2</mn></msup><mo>=</mo><mn>10</mn><mfenced close="]" open="["><mrow><mo>∵</mo><mi>ω</mi><mo>=</mo><mfrac><mi>v</mi><mi>r</mi></mfrac></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo>  </mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mi>m</mi><msup><mi>v</mi><mn>2</mn></msup><mo>=</mo><mn>10</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>=</mo><msqrt><mfrac><mn>40</mn><mrow><mn>3</mn><mi>m</mi></mrow></mfrac></msqrt><mo>=</mo><msqrt><mfrac><mn>40</mn><mrow><mn>3</mn><mo>×</mo><mn>2</mn></mrow></mfrac></msqrt><mo>=</mo><msqrt><mfrac><mn>20</mn><mn>3</mn></mfrac></msqrt><mo> </mo><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p></div>


tags: ['Physics', 'Semiconductors', 'BITSAT', 'BITSAT 2010']
source_db: MarksBatch1_P1.db
title:
a-dc-battery-of-v-volt-is-connected-to-a-series-combination-of-a-resistor-r-and-an-ideal-diode-d-as-shown-in-the-figure-below-the-potential-difference

slug:
a-dc-battery-of-v-volt-is-connected-to-a-series-combination-of-a-resistor-r-and-an-ideal-diode-d-as-shown-in-the-figure-below-the-potential-difference-59496

question:
<div class="question">A d.c. battery of $V$ volt is connected to a series combination of a resistor $R$ and an ideal diode $D$ as shown in the figure below. The potential difference across $R$ will be<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/afB0UvW-f7fUIb0YzgXiL2qyWTZF5Kvk43VjWQhsNW0.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2 V$ when diode is forward biased</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">Zero when diode is forward biased</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$V$ when diode is reverse biased</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$V$ when diode is forward biased</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">Zero when diode is forward biased</span> </div>

answer:
<div class="solution">In forward biasing, the diode conducts. For ideal junction diode, the forward resistance is zero; therefore, entire applied voltage occurs across external resistance R i.e., there occurs no potential drop, so potential across $\mathrm{R}$ is $\mathrm{V}$ in forward biased.</div>


tags: ['Physics', 'Magnetic Effects of Current', 'BITSAT', 'BITSAT 2016']
answer: <div class="solution">No Solution</div>
source_db: MarksBatch1_P1.db
title:
a-deuteron-of-kinetic-energy-50-kev-is-describing-a-circular-orbit-of-radius-05-metre-in-a-plane-perpendicular-to-the-magnetic-field-b-the-kinetic-ene

slug:
a-deuteron-of-kinetic-energy-50-kev-is-describing-a-circular-orbit-of-radius-05-metre-in-a-plane-perpendicular-to-the-magnetic-field-b-the-kinetic-ene-49002

question:
<div class="question">A deuteron of kinetic energy $50 \mathrm{keV}$ is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field $\mathrm{B}$. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same $\mathrm{B}$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$25 \mathrm{keV}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$50 \mathrm{keV}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$200 \mathrm{keV}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$100 \mathrm{keV}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$100 \mathrm{keV}$</span> </div>


tags: ['Mathematics', 'Probability', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-die-is-loaded-such-that-the-probability-of-throwing-the-number-i-is-proportional-to-its-reciprocal-the-probability-that-3-appears-in-a-single-throw-

slug:
a-die-is-loaded-such-that-the-probability-of-throwing-the-number-i-is-proportional-to-its-reciprocal-the-probability-that-3-appears-in-a-single-throw-55951

question:
<div class="question">A die is loaded such that the probability of throwing the number \(i\) is proportional to its reciprocal. The probability that 3 appears in a single throw is-</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">\(3 / 22\)</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\(3 / 11\)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\(9 / 22\)</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">None of these</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">None of these</span> </div>

answer:
<div class="solution">\(\begin{aligned}<br/>&amp; \mathrm{P}(\mathrm{i})=\frac{\mathrm{k}}{\mathrm{i}} \Rightarrow \mathrm{1}=\sum_{\mathrm{i}=1}^6 \mathrm{P}(\mathrm{i})=\mathrm{k} \sum_{\mathrm{i}=1}^6 \frac{1}{\mathrm{i}}=\mathrm{k} \frac{49}{20} \\<br/>&amp; \Rightarrow \mathrm{k}=\frac{20}{49} . \mathrm{P}(3)=\frac{20}{147}<br/>\end{aligned}\)</div>


tags: ['Physics', 'Alternating Current', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-direct-current-of-5-a-is-superposed-on-an-alternating-current-i-10-sin-t-flowing-through-the-wire-the-effective-value-of-the-resulting-current-will-

slug:
a-direct-current-of-5-a-is-superposed-on-an-alternating-current-i-10-sin-t-flowing-through-the-wire-the-effective-value-of-the-resulting-current-will-66409

question:
<div class="question">A direct current of $5 \mathrm{~A}$ is superposed on an alternating current $I=10 \sin \omega t$ flowing through the wire. The effective value of the resulting current will be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$(15 / 2) \mathrm{A}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$5 \sqrt{3} \mathrm{~A}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$5 \sqrt{5} \mathrm{~A}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$15 \mathrm{~A}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$5 \sqrt{3} \mathrm{~A}$</span> </div>

answer:
<div class="solution">Total carrent, $1=(5+10 \sin \omega \mathrm{t})$
<br/> <br/>$$
<br/> <br/>\begin{aligned}
<br/> <br/>\Rightarrow I_{\text {eff }} &amp;=\left[\frac{\int_{0}^{\mathrm{T}} \mathrm{I}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}\right]^{1 / 2} \\
<br/> <br/>&amp;=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(5+10 \sin \omega \mathrm{t})^{2} \mathrm{dt}\right]^{1 / 2} \\
<br/> <br/>&amp;=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}\left(25+100 \sin \omega \mathrm{t}+100 \sin ^{2} \omega \mathrm{t}\right)\right]^{1 / 2} \\
<br/> <br/>\text { But, } \quad \frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin \omega \mathrm{t} \mathrm{dt}=0
<br/> <br/>\end{aligned}
<br/> <br/>$$
<br/> <br/>and
<br/> <br/>$$
<br/> <br/>\begin{array}{l}
<br/> <br/>\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin ^{2} \omega \mathrm{t} \mathrm{dt}=\frac{1}{2} \\
<br/> <br/>\text { So, } \mathrm{I}_{\mathrm{eff}}=\left[25+\frac{1}{2} \times 100\right]^{1 / 2}=5 \sqrt{3} \mathrm{~A}
<br/> <br/>\end{array}
<br/> <br/>$$</div>


tags: ['Physics', 'Waves and Sound', 'BITSAT', 'BITSAT 2022']
source_db: MarksBatch1_P1.db
title:
a-direct-current-of-6-a-is-superimposed-on-an-alternating-current-i-10-sin-t-flowing-through-a-wire-the-effective-value-of-the-resulting-current-will-

slug:
a-direct-current-of-6-a-is-superimposed-on-an-alternating-current-i-10-sin-t-flowing-through-a-wire-the-effective-value-of-the-resulting-current-will-87377

question:
<div class="question">A direct current of $6 \mathrm{~A}$ is superimposed on an alternating current $I=10 \sin \omega t$ flowing through a wire. The effective value of the resulting current will be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"> <br/>$5 \sqrt{2}$
<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"> <br/>$5 \sqrt{3}$
<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"> <br/>$9.27$
<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"> <br/>$8.37$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"> <br/>$9.27$
<br/></span> </div>

answer:
<div class="solution">Given, $I=6+10 \sin \omega t$
<br/>$$
<br/>\begin{aligned}
<br/>I_{\mathrm{eft}} &amp; =\left[\frac{\left.\int_0^T I^2 d t\right]^{1 / 2}}{\int_0^T d t}=\left[\frac{1}{T} \int_0^T(6+10 \sin \omega t)^2 d t\right]^{1 / 2}\right. \\
<br/>&amp; =\left[\frac{1}{T} \int_0^T\left(36+120 \sin \omega t+100 \sin ^2 \omega t\right) d t\right]^{1 / 2}
<br/>\end{aligned}
<br/>$$
<br/>But as, $\frac{1}{T} \int_0^T \sin \omega t d t=0$ and $\frac{1}{T} \int_0^T \sin ^2 \omega t=\frac{1}{2}$
<br/>$$
<br/>\Rightarrow \quad I_{\mathrm{eff}}=\left[36+\frac{1}{2} \times 100\right]^{1 / 2}=9.27
<br/>$$
<br/>Thus, $\quad I_{\text {eff }}=9.27 \mathrm{~A}$.</div>


tags: ['Physics', 'Rotational Motion', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-disc-is-performing-pure-rolling-on-a-smooth-stationary-surface-with-constant-angular-velocity-as-shown-in-figure-at-any-instant-for-the-lower-most-p-1

slug:
a-disc-is-performing-pure-rolling-on-a-smooth-stationary-surface-with-constant-angular-velocity-as-shown-in-figure-at-any-instant-for-the-lower-most-p-1-14701

question:
<div class="question">A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in figure. At any instant, for the lower most point of the disc $-$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/bfu6gxTxLm9omThWIOLYfn6k6jrae-NkE4zEXENd360.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">velocity is v, acceleration is zero</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">velocity is zero, acceleration is
<br/> <br/>zero</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">velocity is v, acceleration is $\mathrm{v}^{2} / \mathrm{R}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">velocity is zero, acceleration is $\mathrm{v}^{2} / \mathrm{R}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">velocity is zero, acceleration is $\mathrm{v}^{2} / \mathrm{R}$</span> </div>

answer:
<div class="solution">As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure $\mathrm{v}_{\text {net }}$ (for lowest point) $=\mathrm{v}-\mathrm{R} \omega=\mathrm{v}-\mathrm{v}=0$
<br/> <br/>and acceleration
<br/> <br/>$=\frac{\mathrm{v}^{2}}{\mathrm{R}}+0=\frac{\mathrm{v}^{2}}{\mathrm{R}}$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/EyZu6lGUGQJRjAdr1OzjBi45khqyYV8BsmqTxCDUKjw.original.fullsize.png"/><br/> <br/> <br/>(since linear speed is constant)</div>


tags: ['Physics', 'Rotational Motion', 'BITSAT', 'BITSAT 2013']
source_db: MarksBatch1_P1.db
title:
a-disc-is-performing-pure-rolling-on-a-smooth-stationary-surface-with-constant-angular-velocity-as-shown-in-figure-at-any-instant-for-the-lower-most-p

slug:
a-disc-is-performing-pure-rolling-on-a-smooth-stationary-surface-with-constant-angular-velocity-as-shown-in-figure-at-any-instant-for-the-lower-most-p-28928


tags: ['Physics', 'Rotational Motion', 'BITSAT', 'BITSAT 2023 (Memory Based Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-disc-is-rolling-without-slipping-on-a-surface-the-radius-of-the-disc-is-r-at-t-0-the-top-most-point-on-the-disc-is-a-as-shown-in-figure-when-the-dis

slug:
a-disc-is-rolling-without-slipping-on-a-surface-the-radius-of-the-disc-is-r-at-t-0-the-top-most-point-on-the-disc-is-a-as-shown-in-figure-when-the-dis-47624

question:
<div class="question">A disc is rolling without slipping on a surface. The radius of the disc is $R$. At $t=0$, the top most point on the disc is $\mathrm{A}$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/q-SlnnL2YNKWPjSCnEgbYs-9BTAVQuKIBpDRx5eQUoE.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\mathrm{R} \sqrt{\left(\pi^2+4\right)}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{R} \sqrt{\left(\pi^2+1\right)}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$2 \mathrm{R}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$2 \mathrm{R} \sqrt{\left(1+4 \pi^2\right)}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{R} \sqrt{\left(\pi^2+4\right)}$</span> </div>

answer:
<div class="solution">(a) From figure,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/Djr7jOTnwC9j_KyyquLbu_JqUqhoVr-OQSeyH17AjIk.original.fullsize.png"/><br/><br/>Displacement,<br/>$\mathrm{BA}=\sqrt{(2 \mathrm{R})^2+(\pi \mathrm{R})^2}=\mathrm{R} \sqrt{4+\pi^2}$</div>


tags: ['Physics', 'Kinetic Theory of Gases', 'BITSAT', 'BITSAT 2023 (Memory Based Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-flask-contains-hydrogen-and-oxygen-in-the-ratio-of-2-1-by-mass-at-temperature-2-7-c-the-ratio-of-average-kinetic-energy-per-molecule-of-hydrogen-and

slug:
a-flask-contains-hydrogen-and-oxygen-in-the-ratio-of-2-1-by-mass-at-temperature-2-7-c-the-ratio-of-average-kinetic-energy-per-molecule-of-hydrogen-and-41431

question:
<div class="question">A flask contains hydrogen and oxygen in the ratio of $2: 1$ by mass at temperature $27^{\circ} \mathrm{C}$. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is :</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>2 : 1</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>1 : 1</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>1 : 4</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>4 : 1</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>1 : 1</span> </div>

answer:
<div class="solution">Average kinetic energy for diatomic gases<br/>$\begin{aligned}<br/>&amp; \mathrm{K}_{\mathrm{av}}=\frac{5}{2} \mathrm{kT} \\<br/>&amp; \therefore \frac{\left(\mathrm{K}_{\mathrm{av}}\right)_{\mathrm{H}}}{\left(\mathrm{K}_{\mathrm{av}}\right)_{\mathrm{O}}}=\frac{(27+273)}{(27+273)}=1<br/>\end{aligned}$</div>


tags: ['Physics', 'Laws of Motion', 'BITSAT', 'BITSAT 2016']
source_db: MarksBatch1_P1.db
title:
a-frictionless-wire-ab-is-fixed-on-a-sphere-of-radius-r-a-very-small-spherical-ball-slips-on-this-wire-the-time-taken-by-this-ball-to-slip-from-a-to-b

slug:
a-frictionless-wire-ab-is-fixed-on-a-sphere-of-radius-r-a-very-small-spherical-ball-slips-on-this-wire-the-time-taken-by-this-ball-to-slip-from-a-to-b-87941

question:
<div class="question">A frictionless wire $\mathrm{AB}$ is fixed on a sphere of radius $\mathrm{R}$. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to $\mathrm{B}$ is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/r_OrMPvilqMNy6lx1g52JXylmmTCE4Iq5qfXm1-nk1k.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{\sqrt{2 g R}}{g \cos \theta}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$2 \sqrt{g R} \cdot \frac{\cos \theta}{g}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$2 \sqrt{\frac{R}{g}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{g R}{\sqrt{g \cos \theta}}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$2 \sqrt{\frac{R}{g}}$</span> </div>

answer:
<div class="solution">Acceleration of body along $\mathrm{AB}$ is g $\cos \theta$ Distance travelled in time $\mathrm{t}$ sec $=$
<br/> <br/>$A B=\frac{1}{2}(g \cos \theta) t^{2}$
<br/> <br/>From $\Delta \mathrm{ABC}, \mathrm{AB}=2 \mathrm{R} \cos \theta$
<br/> <br/>Thus, $2 \mathrm{R} \cos \theta=\frac{1}{2} g \cos \theta t^{2}$
<br/> <br/>$\Rightarrow t^{2}=\frac{4 R}{g} \Rightarrow t=2 \sqrt{\frac{R}{g}}$</div>


tags: ['Physics', 'Kinetic Theory of Gases', 'BITSAT', 'BITSAT 2018']
source_db: MarksBatch1_P1.db
title:
a-gas-has-molar-heat-capacity-c-37-55-j-mol-1-k-1-in-this-process-p-t-constant-the-number-of-degree-of-freedom-of-the-molecule-of-gas-is

slug:
a-gas-has-molar-heat-capacity-c-37-55-j-mol-1-k-1-in-this-process-p-t-constant-the-number-of-degree-of-freedom-of-the-molecule-of-gas-is-97143

question:
<div class="question">A gas has molar heat capacity <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>=</mo><mn>37</mn><mo>.</mo><mn>55</mn><mo> </mo><mi mathvariant="normal">J</mi><mo> </mo><msup><mi>mol</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mo> </mo><msup><mi mathvariant="normal">K</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math>, in this process <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>p</mi><mi>T</mi><mo>=</mo></math> constant. The number of degree of freedom of the molecule of gas is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>5</mn></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>7</mn></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>5</mn></math></span> </div>

answer:
<div class="solution"><p>Given, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>=</mo><mn>37</mn><mo>.</mo><mn>55</mn><mo> </mo><mi mathvariant="normal">J</mi><mo> </mo><msup><mi>mol</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mo> </mo><msup><mi mathvariant="normal">K</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p><p>Also, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>T</mi><mo>=</mo></math> constant <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi></math></p><p>According to ideal gas equation, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>V</mi><mo>=</mo><mi>R</mi><mi>T</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mo> </mo><mo> </mo><mi>p</mi><mo>=</mo><mfrac><mrow><mi>R</mi><mi>T</mi></mrow><mi>V</mi></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mi mathvariant="normal">i</mi></mfenced></math></p><p>Putting the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>p</mi></math> in Eq. (i), we get</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>R</mi><mi>T</mi></mrow><mi>V</mi></mfrac><mo>×</mo><mi>T</mi><mo>=</mo><mi>K</mi><mo> </mo><mo> </mo><mo> </mo><mo>⇒</mo><mo> </mo><mo> </mo><mo> </mo><mi>V</mi><mo>=</mo><mfrac><mrow><mi>R</mi><msup><mi>T</mi><mn>2</mn></msup></mrow><mi>K</mi></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mi>ii</mi></mfenced></math></p><p>On differentiating above equation both sides, we get</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>d</mi><mi>V</mi></mrow><mrow><mi>d</mi><mi>T</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>2</mn><mi>R</mi><mi>T</mi></mrow><mi>K</mi></mfrac><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mi>iii</mi></mfenced></math><br/>But <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>T</mi><mi>K</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><mi>p</mi></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mfenced close="]" open="["><mrow><mi>from</mi><mi>Eq</mi><mo>.</mo><mo> </mo><mo> </mo><mfenced><mi mathvariant="normal">i</mi></mfenced></mrow></mfenced></math><br/>Hence, Eq. <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mi>iii</mi></mfenced></math>becomes</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>d</mi><mi>V</mi></mrow><mrow><mi>d</mi><mi>T</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>2</mn><mo> </mo><mi mathvariant="normal">R</mi></mrow><mi>p</mi></mfrac></math><br/>So, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>=</mo><msub><mi>C</mi><mi>V</mi></msub><mo>+</mo><mfrac><mrow><mi>p</mi><mi>d</mi><mi>V</mi></mrow><mrow><mi>d</mi><mi>T</mi></mrow></mfrac></math><br/>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>=</mo><msub><mi>C</mi><mi>V</mi></msub><mo>+</mo><mfrac><mrow><mi>p</mi><mo>×</mo><mn>2</mn><mo> </mo><mi mathvariant="normal">R</mi></mrow><mi>p</mi></mfrac><mo>=</mo><msub><mi>C</mi><mi>V</mi></msub><mo>+</mo><mn>2</mn><mo> </mo><mi mathvariant="normal">R</mi></math><br/>or <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mi>V</mi></msub><mo>=</mo><mi>C</mi><mo>-</mo><mn>2</mn><mo> </mo><mi mathvariant="normal">R</mi><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mi>iv</mi></mfenced></math></p><p>As <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mi>V</mi></msub><mo>=</mo><mfrac><mrow><mi>n</mi><mi>R</mi></mrow><mn>2</mn></mfrac><mo>,</mo></math> where <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>n</mi><mo>=</mo></math> number of degrees of freedom.<br/>Putting the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mi>V</mi></msub></math> in Eq. (iv), we get</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>n</mi><mi>R</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mi>C</mi><mo>-</mo><mn>2</mn><mo> </mo><mi mathvariant="normal">R</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>n</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mo> </mo><mfenced><mrow><mi>C</mi><mo>-</mo><mn>2</mn><mo> </mo><mi mathvariant="normal">R</mi></mrow></mfenced></mrow><mi>R</mi></mfrac><mo>=</mo><mfrac><mrow><mn>2</mn><mo> </mo><mfenced><mrow><mn>37</mn><mo>.</mo><mn>55</mn><mo>-</mo><mn>2</mn><mo>×</mo><mn>8</mn><mo>.</mo><mn>3</mn></mrow></mfenced></mrow><mrow><mn>8</mn><mo>.</mo><mn>3</mn></mrow></mfrac><mo>=</mo><mn>5</mn><mo>.</mo><mn>048</mn><mo>≈</mo><mn>5</mn></math></p></div>


tags: ['Chemistry', 'Redox Reactions', 'BITSAT', 'BITSAT 2011']
answer: <div class="solution">The more the reduction potential, the more the oxidising power.</div>
source_db: MarksBatch1_P1.db
title:
a-gas-x-at-1-atm-is-bubbled-through-a-solution-containing-a-mixture-of-1-m-y-and-1-m-z-at-2-5-c-if-the-reduction-potential-is-z-y-x-then

slug:
a-gas-x-at-1-atm-is-bubbled-through-a-solution-containing-a-mixture-of-1-m-y-and-1-m-z-at-2-5-c-if-the-reduction-potential-is-z-y-x-then-18267

question:
<div class="question">A gas $X$ at 1 atm is bubbled through a solution containing a mixture of $1 \mathrm{M} \mathrm{Y}^{-}$ and $1 \mathrm{M} \mathrm{Z}^{-}$ at $25^{\circ} \mathrm{C}$. If the reduction potential is $Z&gt;Y&gt;X$, then</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\mathrm{Y}$ will oxidise $\mathrm{X}$ and $\mathrm{not} \mathrm{Z}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{Y}$ will oxidise $Z$ and $\mathrm{not} \mathrm{X}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{Y}$ will oxidise both $\mathrm{X}$ and $\mathrm{Z}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{Y}$ will reduce both $\mathrm{X}$ and $\mathrm{Z}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{Y}$ will oxidise $\mathrm{X}$ and $\mathrm{not} \mathrm{Z}$</span> </div>


tags: ['Chemistry', 'Chemical Kinetics', 'BITSAT', 'BITSAT 2014']
source_db: MarksBatch1_P1.db
title:
a-gaseous-reaction-x-2-g-y-2-1-z-g-there-is-increase-in-pressure-from-100-mm-to-120-mm-in-5-minutes-the-rate-of-disappearance-of-x-2-is

slug:
a-gaseous-reaction-x-2-g-y-2-1-z-g-there-is-increase-in-pressure-from-100-mm-to-120-mm-in-5-minutes-the-rate-of-disappearance-of-x-2-is-84137

question:
<div class="question">A gaseous reaction $\mathrm{X}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Y}+\frac{1}{2} \mathrm{Z}(\mathrm{g})$
<br/> <br/>There is increase in pressure from $100 \mathrm{~mm}$ to 120 $\mathrm{mm}$ in 5 minutes. The rate of disappearance of $\mathrm{X}_{2}$ is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$8 \mathrm{~mm} \mathrm{~min}^{-1}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$2 \mathrm{~mm} \mathrm{~min}^{-1}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$16 \mathrm{~mm} \mathrm{~min}^{-1}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$4 \mathrm{~mm} \min ^{-1}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$8 \mathrm{~mm} \mathrm{~min}^{-1}$</span> </div>

answer:
<div class="solution">The increase in pressure shows the increase in conc. of $Z$. Rate of appearance of $Z=$
<br/> <br/>$$
<br/> <br/>\frac{120-100}{5}=4 \mathrm{~mm} \mathrm{~min}^{-1}
<br/> <br/>$$
<br/> <br/>Rate of disappearance of $\mathrm{X}_{2}=2 \times$ rate of appearance of $Z$ $=2 \times 4 \mathrm{~mm} \mathrm{~min}^{-1}=8 \mathrm{~mm} \mathrm{~min}^{-1}$</div>


tags: ['Physics', 'Gravitation', 'BITSAT', 'BITSAT 2017']
source_db: MarksBatch1_P1.db
title:
a-geostationary-satellite-orbits-around-the-earth-in-a-circular-orbit-of-radius-36000-km-then-the-time-period-of-a-sky-satellite-orbiting-a-few-100-km

slug:
a-geostationary-satellite-orbits-around-the-earth-in-a-circular-orbit-of-radius-36000-km-then-the-time-period-of-a-sky-satellite-orbiting-a-few-100-km-46380

question:
<div class="question">A geostationary satellite orbits around the earth in a circular orbit of radius <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>36000</mn><mo> </mo><mi>km</mi></math>. Then, the time period of a sky satellite orbiting a few <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>100</mn><mo> </mo><mi>km</mi></math> above the earth's surface <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mi>R</mi><mo>=</mo><mn>64000</mn><mo> </mo><mi>km</mi></mrow></mfenced></math> will approximately be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac><mo> </mo><mi mathvariant="normal">h</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo> </mo><mi mathvariant="normal">h</mi></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo> </mo><mi mathvariant="normal">h</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mo> </mo><mi mathvariant="normal">h</mi></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo> </mo><mi mathvariant="normal">h</mi></math></span> </div>

answer:
<div class="solution"><p>By Kepler's law of planetary motion,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>T</mi><mn>2</mn></msup><mo>∝</mo><msup><mi>r</mi><mn>3</mn></msup></math>, hence <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi>T</mi><mn>1</mn><mn>2</mn></msubsup><mo>∝</mo><msubsup><mi>r</mi><mn>1</mn><mn>3</mn></msubsup></math></p><p>Similarly <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi>T</mi><mn>2</mn><mn>2</mn></msubsup><mo>∝</mo><msubsup><mi>r</mi><mn>2</mn><mn>3</mn></msubsup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><msubsup><mi>T</mi><mn>1</mn><mn>2</mn></msubsup><mo>∝</mo><msup><mfenced><mn>36000</mn></mfenced><mn>3</mn></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi>T</mi><mn>2</mn><mn>2</mn></msubsup><mo>∝</mo><msup><mfenced><mrow><mn>6400</mn><mo>+</mo><mi>h</mi></mrow></mfenced><mn>3</mn></msup></math></p><p>Therefore, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><msub><mi>T</mi><mn>2</mn></msub></mfenced><mn>2</mn></msup><mo>=</mo><msubsup><mi>T</mi><mn>1</mn><mn>2</mn></msubsup><msup><mfenced><mfrac><mrow><mn>6400</mn><mo>+</mo><mi>h</mi></mrow><mn>36000</mn></mfrac></mfenced><mn>3</mn></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&gt;</mo><msubsup><mi>T</mi><mn>1</mn><mn>2</mn></msubsup><msup><mfenced close="]" open="["><mfrac><mn>6400</mn><mn>3600</mn></mfrac></mfenced><mn>3</mn></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&gt;</mo><msup><mfenced><mn>24</mn></mfenced><mn>2</mn></msup><msup><mfenced close="]" open="["><mfrac><mn>8</mn><mn>45</mn></mfrac></mfenced><mn>3</mn></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><msub><mi>T</mi><mn>2</mn></msub><mo>&gt;</mo><mfrac><mrow><mn>24</mn><mo>×</mo><mn>8</mn><mo>×</mo><msqrt><mn>8</mn></msqrt></mrow><mrow><mn>45</mn><mo>×</mo><msqrt><mn>45</mn></msqrt></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&gt;</mo><mn>1</mn><mo>.</mo><mn>8</mn><mo> </mo><mi mathvariant="normal">h</mi></math></p><p>So, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>T</mi><mn>2</mn></msub><mo>≅</mo><mn>2</mn><mo> </mo><mi mathvariant="normal">h</mi></math></p></div>


tags: ['Physics', 'Mechanical Properties of Fluids', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-glass-capillary-tube-of-internal-radius-r-025-mm-is-immersed-in-water-the-top-end-of-the-tube-projected-by-2-cm-above-the-surface-of-the-water-at-wh

slug:
a-glass-capillary-tube-of-internal-radius-r-025-mm-is-immersed-in-water-the-top-end-of-the-tube-projected-by-2-cm-above-the-surface-of-the-water-at-wh-25796

question:
<div class="question">A glass capillary tube of internal radius $\mathrm{r}=0.25$ $\mathrm{mm}$ is immersed in water. The top end of the tube projected by $2 \mathrm{~cm}$ above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water $=0.7 \mathrm{~N} / \mathrm{m}$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\theta=90^{\circ}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\quad \theta=70^{\circ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\theta=45^{\circ}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\theta=35^{\circ}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\quad \theta=70^{\circ}$</span> </div>

answer:
<div class="solution">Water wets glass and so the angle of contact is zero. For full rise, neglecting the small mass in the meniscus
<br/> <br/>$$
<br/> <br/>\begin{aligned}
<br/> <br/>2 \pi \mathrm{T} &amp;=\pi \mathrm{r}^{2} \mathrm{~h} \rho \mathrm{g} \Rightarrow \mathrm{h}=\frac{2 \mathrm{~T}}{\mathrm{r} \rho \mathrm{g}}\left[\because \text { water wets glass, } \theta=0^{\circ}\right] \\
<br/> <br/>&amp;=\frac{2 \times 007}{0.25 \times 10^{-3} \times 1000 \times 9.8}
<br/> <br/>\end{aligned}
<br/> <br/>$$
<br/> <br/>As the tube is only $2 \mathrm{~cm}$ above the water and so, water will rise by $2 \mathrm{~cm}$ and meet the tube at an angle such that,
<br/> <br/>$$
<br/> <br/>\begin{aligned}
<br/> <br/>2 \pi r &amp; \operatorname{T} \cos \theta=\pi r^{2} h^{\prime} \rho g \\
<br/> <br/>\Rightarrow &amp; 2 T \cos \theta=h^{\prime} r \rho g \\
<br/> <br/>\Rightarrow \cos \theta &amp;=\frac{h^{\prime} r \rho g}{2 T} \\
<br/> <br/>&amp;=\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}
<br/> <br/>\end{aligned}
<br/> <br/>$$
<br/> <br/>The liquid will meet the tube at an angle, $\theta$ $\cong 70^{\circ}$</div>


tags: ['Chemistry', 'Chemical Kinetics', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-graph-is-plotted-between-log-k-versus-1-t-for-calculation-of-activation-energy-e-a-the-correct-plot-is-1

slug:
a-graph-is-plotted-between-log-k-versus-1-t-for-calculation-of-activation-energy-e-a-the-correct-plot-is-1-25475

question:
<div class="question"><p>A graph is plotted between log <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">K</mi></math> versus <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mi mathvariant="normal">T</mi></mfrac></math> for calculation of activation energy <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><msub><mi mathvariant="normal">E</mi><mi mathvariant="normal">a</mi></msub></mfenced></math>. The correct plot is</p></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/95bf569a-d489-4f23-879f-7c451762bb6a-image.png" style="width: 200px; height: 180px;"/></p></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/dc277033-e689-451f-957f-db816643eb9f-image.png" style="width: 200px; height: 185px;"/></p></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/c8fcacc6-93ca-4780-839b-b5d0c06302c5-image.png" style="width: 200px; height: 186px;"/></p></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/ac619cb0-46a5-4bbf-b5af-c408f8e2774f-image.png" style="width: 200px; height: 161px;"/></p></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/dc277033-e689-451f-957f-db816643eb9f-image.png" style="width: 200px; height: 185px;"/></p></span> </div>

answer:
<div class="solution"><p>According to Arrhenius equation,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">k</mi><mo>∝</mo><mi mathvariant="normal">T</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">k</mi><mo>=</mo><msup><mi>Ae</mi><mrow><mo>−</mo><mi>Ea</mi><mo>​</mo><mo>/</mo><mi>RT</mi></mrow></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>logk</mi><mo>=</mo><mi>logA</mi><mo>-</mo><mfrac><msub><mi mathvariant="normal">E</mi><mi mathvariant="normal">a</mi></msub><mrow><mn>2</mn><mo>.</mo><mn>303</mn><mo>×</mo><mi mathvariant="normal">R</mi></mrow></mfrac><mo>​</mo><mo>​</mo><mo>×</mo><mfrac><mn>1</mn><mi mathvariant="normal">T</mi></mfrac><mo>​</mo></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">y</mi><mo>=</mo><mi>mx</mi><mo>+</mo><mi mathvariant="normal">c</mi></math>, the equation of straight line</p><p>The value of log <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">k</mi></math> should increase uniformily with <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">T</mi></math> or decrease with <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mi mathvariant="normal">T</mi></mfrac></math>.</p><p>               <img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/dc277033-e689-451f-957f-db816643eb9f-image.png" style="width: 200px; height: 185px;"/></p><p>From this, that the slope is negative, which means that angle with x - axis is greater than 90° also that the point on which the straight line of graph intercepts the y - axis is log A which is positive.</p><p>Hence all the conclusions point out towards option b as it has negative slope and a positive y intercept.</p></div>


tags: ['Chemistry', 'Chemical Kinetics', 'BITSAT', 'BITSAT 2017']
source_db: MarksBatch1_P1.db
title:
a-graph-is-plotted-between-log-k-versus-1-t-for-calculation-of-activation-energy-e-a-the-correct-plot-is

slug:
a-graph-is-plotted-between-log-k-versus-1-t-for-calculation-of-activation-energy-e-a-the-correct-plot-is-55239


tags: ['Physics', 'Motion In Two Dimensions', 'BITSAT', 'BITSAT 2021']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">$3: 1$</span> </div>
source_db: MarksBatch1_P1.db
title:
a-gun-fires-two-bullets-at-6-0-and-3-0-with-horizontal-the-bullets-strike-at-some-horizontal-distance-the-ratio-of-maximum-height-for-the-two-bullets-

slug:
a-gun-fires-two-bullets-at-6-0-and-3-0-with-horizontal-the-bullets-strike-at-some-horizontal-distance-the-ratio-of-maximum-height-for-the-two-bullets-87775

question:
<div class="question">A gun fires two bullets at $60^{\circ}$ and $30^{\circ}$ with horizontal. The bullets strike at some horizontal distance. The ratio of maximum height for the two bullets is in the ratio of</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2: 1$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$3: 1$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$4: 1$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$1: 1$</span> </li> </ul>

answer:
<div class="solution">The bullets are fired at the same initial speed
<br/> <br/>$\begin{array}{l}
<br/> <br/>\frac{\mathrm{H}}{\mathrm{H}^{\prime}}=\frac{\mathrm{u}^{2} \sin ^{2} 60^{\circ}}{2 \mathrm{~g}} \times \frac{2 \mathrm{~g}}{\mathrm{u}^{2} \sin ^{2} 30^{\circ}}=\frac{\sin ^{2} 60^{\circ}}{\sin ^{2} 30^{\circ}} \\
<br/> <br/>=\frac{(\sqrt{3} / 2)^{2}}{(1 / 2)^{2}}=\frac{3}{1}
<br/> <br/>\end{array}$</div>


tags: ['Physics', 'Electrostatics', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-hollow-metal-sphere-of-radius-5-cm-is-charged-such-that-the-potential-on-its-surface-is-10-v-the-potential-at-a-distance-of-2-cm-from-the-centre-of-

slug:
a-hollow-metal-sphere-of-radius-5-cm-is-charged-such-that-the-potential-on-its-surface-is-10-v-the-potential-at-a-distance-of-2-cm-from-the-centre-of-21318

question:
<div class="question">A hollow metal sphere of radius $5 \mathrm{~cm}$ is charged such that the potential on its surface is $10 \mathrm{~V}$. The potential at a distance of $2 \mathrm{~cm}$ from the centre of the sphere is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">zero</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$10 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$4 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$10 / 3 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$10 \mathrm{~V}$</span> </div>

answer:
<div class="solution">Potential at any point inside the sphere $=$ potential at the surface of the sphere $=10 \mathrm{~V}$.</div>


tags: ['Physics', 'Magnetic Effects of Current', 'BITSAT', 'BITSAT 2012']
source_db: MarksBatch1_P1.db
title:
a-horizontal-overhead-powerline-is-at-height-of-4-m-from-the-ground-and-carries-a-current-of-100-a-from-east-to-west-the-magnetic-field-directly-below

slug:
a-horizontal-overhead-powerline-is-at-height-of-4-m-from-the-ground-and-carries-a-current-of-100-a-from-east-to-west-the-magnetic-field-directly-below-63585

question:
<div class="question">A horizontal overhead powerline is at height of $4 \mathrm{~m}$ from the ground and carries a current of $100 \mathrm{~A}$ from east to west. The magnetic field directly below it on the ground is $\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\right)$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2.5 \times 10^{-7} \mathrm{~T}$ southward</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$5 \times 10^{-6}$ T northward</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$5 \times 10^{-6} \mathrm{~T}$ southward</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$2.5 \times 10^{-7}$ T northward</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$5 \times 10^{-6} \mathrm{~T}$ southward</span> </div>

answer:
<div class="solution">The magnetic field is
<br/>$\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}}{\mathrm{r}}=10^{-7} \times \frac{2 \times 100}{4}=5 \times 10^{-6} \mathrm{~T}$
<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/fXNqXasjtRkq7PU3M79o92MfBcmfgd2vd8nB1h7sNF4.original.fullsize.png"/><br/> <br/>According to right hand palm rule, the magnetic field is directed towards south.</div>


tags: ['Mathematics', 'Heights and Distances', 'BITSAT', 'BITSAT 2022']
source_db: MarksBatch1_P1.db
title:
a-house-subtends-a-right-angle-at-the-window-of-a-opposite-house-and-the-angle-of-elevation-of-the-window-from-the-bottom-of-the-first-house-is-60-if-

slug:
a-house-subtends-a-right-angle-at-the-window-of-a-opposite-house-and-the-angle-of-elevation-of-the-window-from-the-bottom-of-the-first-house-is-60-if-66761

question:
<div class="question">A house subtends a right angle at the window of a opposite house and the angle of elevation of the window from the bottom of the first house is $60$. If the distance between two houses be $6 \mathrm{~m}$, then the height of the first house is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"> <br/>$8 \sqrt{3} \mathrm{~m}$
<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"> <br/>$6 \sqrt{3} \mathrm{~m}$
<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"> <br/>$4 \sqrt{3} \mathrm{~m}$
<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"> <br/>None of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"> <br/>$8 \sqrt{3} \mathrm{~m}$
<br/></span> </div>

answer:
<div class="solution">Let $P Q$ be the house subtending a right angle at the window $B$ of opposite house $A B$
<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/0V7dsPYHXJPSlfmOThYsY38RGsVYM4qnm7QqqLpWp0s.original.fullsize.png"/><br/> <br/>In $\triangle P A B$, we have 
<br/>$\tan 60^{\circ}=\frac{A B}{6}=A B=6 \sqrt{3} \mathrm{~m}$
<br/>In $\triangle C B Q$, we have
<br/>$\tan 30^{\circ}=\frac{h-C P}{B C}$
<br/>$\begin{array}{ll}\Rightarrow &amp; \frac{1}{\sqrt{3}}=\frac{h-6 \sqrt{3}}{6} \quad[\because A B=C P, B C=A P] \\ \Rightarrow &amp; h=6\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right) \\ \Rightarrow &amp; h=8 \sqrt{3} \mathrm{~m}\end{array}$</div>


tags: ['Physics', 'Thermodynamics', 'BITSAT', 'BITSAT 2023 (Memory Based Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-hypothetical-gas-expands-adiabatically-such-that-its-volume-changes-from-08-litres-to-27-litres-if-the-ratio-of-final-pressure-of-the-gas-to-initial

slug:
a-hypothetical-gas-expands-adiabatically-such-that-its-volume-changes-from-08-litres-to-27-litres-if-the-ratio-of-final-pressure-of-the-gas-to-initial-32337

question:
<div class="question">A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$. Then the ratio of $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}$ will be.</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\frac{4}{3}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{3}{1}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{1}{2}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{3}{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{4}{3}$</span> </div>

answer:
<div class="solution">In adiabatic process<br/>$\begin{aligned}<br/>&amp; \mathrm{PV}^\gamma=\text { constant } \\<br/>&amp; \Rightarrow \quad \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma \\<br/>&amp; \Rightarrow \quad \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^\gamma \Rightarrow \frac{81}{16}=\left(\frac{27}{8}\right)^\gamma \\<br/>&amp; \Rightarrow\left(\frac{9}{4}\right)^2=\left(\frac{9 \times 3}{4 \times 2}\right)^\gamma \Rightarrow\left(\frac{9}{4}\right)^{2-\gamma}=\left(\frac{3}{2}\right)^\gamma \\<br/>&amp; \Rightarrow\left(\frac{9}{4}\right)^{2-\gamma}=\left(\frac{9}{4}\right)^{\gamma / 2} \Rightarrow 2-\gamma=\frac{\gamma}{2} \\<br/>&amp; \Rightarrow 2=\frac{3}{2} \gamma \Rightarrow \gamma=\frac{4}{3}<br/>\end{aligned}$</div>


tags: ['Chemistry', 'Hydrocarbons', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-is-an-optically-inactive-alkyl-chloride-which-on-reaction-with-aqueous-koh-gives-b-b-on-heating-with-cu-at-30-0-c-gives-an-alkene-c-what-are-a-and-c-1

slug:
a-is-an-optically-inactive-alkyl-chloride-which-on-reaction-with-aqueous-koh-gives-b-b-on-heating-with-cu-at-30-0-c-gives-an-alkene-c-what-are-a-and-c-1-36735

question:
<div class="question">A is an optically inactive alkyl chloride which on reaction with aqueous $\mathrm{KOH}$ gives $\mathrm{B}$. $\mathrm{B}$ on heating with $\mathrm{Cu}$ at $300^{\circ} \mathrm{C}$ gives an alkene $\mathrm{C}$, what are $\mathrm{A}$ and $\mathrm{C}$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{CH}_{2}=\mathrm{CH}_{2}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{MeCH}=\mathrm{CH} \cdot \mathrm{Me}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{Me}_{2} \mathrm{CH} . \mathrm{CH}_{2} \mathrm{Cl} \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$</span> </div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/r_jzUIv-Nw7CGGxbT1JMECZ_8-bXs_RLrzK7s_1eGWM.original.fullsize.png"/><br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/A1h6mVXZ-ohdv4SfRu_QUNMXYUlgXa4q03B3lnVhDZI.original.fullsize.png"/><br/><br/>OR<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/aOvKvW1brEWLAWQwhcMdHf0OfbXvuvTVQE1gw4h2F5Y.original.fullsize.png"/></div>


tags: ['Chemistry', 'Hydrocarbons', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-is-an-optically-inactive-alkyl-chloride-which-on-reaction-with-aqueous-koh-gives-b-b-on-heating-with-cu-at-30-0-c-gives-an-alkene-c-what-are-a-and-c

slug:
a-is-an-optically-inactive-alkyl-chloride-which-on-reaction-with-aqueous-koh-gives-b-b-on-heating-with-cu-at-30-0-c-gives-an-alkene-c-what-are-a-and-c-35466


tags: ['Mathematics', 'Heights and Distances', 'BITSAT', 'BITSAT 2011']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">12</span> </div>
answer: <div class="solution">Length of ladder \(=\frac{6 \sqrt{3}}{\sin 60^{\circ}}=12 \mathrm{~m}\)</div>
source_db: MarksBatch1_P1.db
title:
a-ladder-rests-against-a-wall-so-that-its-top-touches-the-roof-of-the-house-if-the-ladder-makes-an-angle-of-60circ-with-the-horizontal-and-height-of-t

slug:
a-ladder-rests-against-a-wall-so-that-its-top-touches-the-roof-of-the-house-if-the-ladder-makes-an-angle-of-60circ-with-the-horizontal-and-height-of-t-99051

question:
<div class="question">A ladder rests against a wall so that its top touches the roof of the house. If the ladder makes an angle of \(60^{\circ}\) with the horizontal and height of the house be \(6 \sqrt{3}\) meters, then the length of the ladder in meters is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">\(12 \sqrt{3}\)</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">12</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\(12 / \sqrt{3}\)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">None of these</span> </li> </ul>


tags: ['Physics', 'Mechanical Properties of Fluids', 'BITSAT', 'BITSAT 2012']
source_db: MarksBatch1_P1.db
title:
a-large-drop-of-oil-density-08-g-cm-3-and-viscosity-0-floats-up-through-a-column-of-another-liquid-density-12-g-cm-3-and-viscosity-l-assuming-that-the

slug:
a-large-drop-of-oil-density-08-g-cm-3-and-viscosity-0-floats-up-through-a-column-of-another-liquid-density-12-g-cm-3-and-viscosity-l-assuming-that-the-66066

question:
<div class="question">A large drop of oil (density $0.8 \mathrm{~g} / \mathrm{cm}^{3}$ and viscosity $\eta_{0}$ ) floats up through a column of another liquid (density $1.2 \mathrm{~g} / \mathrm{cm}^{3}$ and viscosity $\eta_{\mathrm{L}}$ ). Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on:</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\eta_{0}$ only</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\eta_{L}$ only</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">both on $\eta_{0}$ and $\eta_{\mathrm{L}}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">neither $\eta_{0}$ nor $\eta_{L}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\eta_{L}$ only</span> </div>

answer:
<div class="solution">Forces of viscosity arise in a fluid when different sections of the fluid move with different velocities. Here, there is no such motion inside the oil drop and hence the viscosity of oil does not come into play.</div>


tags: ['Physics', 'Ray Optics', 'BITSAT', 'BITSAT 2016']
source_db: MarksBatch1_P1.db
title:
a-lens-having-focal-length-f-and-aperture-of-diameter-d-forms-an-image-of-intensity-i-aperture-of-diameter-2-d-in-central-region-of-lens-is-covered-by

slug:
a-lens-having-focal-length-f-and-aperture-of-diameter-d-forms-an-image-of-intensity-i-aperture-of-diameter-2-d-in-central-region-of-lens-is-covered-by-75234

question:
<div class="question">A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$f$ and $\frac{I}{4}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{3 f}{4}$ and $\frac{I}{2}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$f$ and $\frac{3 I}{4}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{f}{2}$ and $\frac{I}{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$f$ and $\frac{3 I}{4}$</span> </div>

answer:
<div class="solution">Bycovering aperture, focal length does not change. But intensity is reduced by $\frac{1}{4}$ times
<br/> <br/>as aperture diameter $\frac{d}{2}$ is covered.
<br/> <br/>$$
<br/> <br/>\begin{array}{l}
<br/> <br/>\therefore \quad I^{\prime}=I-\frac{I}{4}=\frac{3 I}{4} \\
<br/> <br/>\therefore \quad \text { New focal length }=\mathrm{f} \text { and intensity }=\frac{3 I}{4} .
<br/> <br/>\end{array}
<br/> <br/>$$</div>


tags: ['Physics', 'Current Electricity', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-lettera-is-constructed-of-a-uniform-wire-with-resistance-10-per-cm-the-sides-of-the-letter-are-20-cm-and-the-cross-piece-in-the-middle-is-10-cm-long-1

slug:
a-lettera-is-constructed-of-a-uniform-wire-with-resistance-10-per-cm-the-sides-of-the-letter-are-20-cm-and-the-cross-piece-in-the-middle-is-10-cm-long-1-91618

question:
<div class="question">A letter'A' is constructed of a uniform wire with resistance $1.0 \Omega$ per $\mathrm{cm}$, The sides of the letter are $20 \mathrm{~cm}$ and the cross piece in the middle is 10 $\mathrm{cm}$ long. The apex angle is $60 .$ The resistance between the ends of the legs is close to:</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$50.0 \Omega$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$10 \Omega$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$36.7 \Omega$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$26.7 \Omega$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$26.7 \Omega$</span> </div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/Y5sbUODQvszObE9Pnn_44LN51U94r0ntiNbllSp34b8.original.fullsize.png"/><br/> <br/>For ADE $\frac{1}{R^{\prime}}=\frac{1}{2 x}+\frac{1}{10} \quad$ or $\quad R^{\prime}=\frac{20 x}{10+2 x}$
<br/>$\mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x} \quad \ldots$(i)
<br/>or $\quad \frac{20 \mathrm{x}}{10+2 \mathrm{x}}+40=2 \mathrm{x}$
<br/>Solving we get $\mathrm{x}=10 \Omega$
<br/>Putting the value of $x=10 \Omega$ in equation (i) We get $\mathrm{R}_{\mathrm{BC}}=\frac{20 \times 10}{10+2 \times 10}+20-10+20-10$
<br/>$=\frac{80}{3}=26.7 \Omega$</div>


tags: ['Physics', 'Current Electricity', 'BITSAT', 'BITSAT 2012']
source_db: MarksBatch1_P1.db
title:
a-lettera-is-constructed-of-a-uniform-wire-with-resistance-10-per-cm-the-sides-of-the-letter-are-20-cm-and-the-cross-piece-in-the-middle-is-10-cm-long

slug:
a-lettera-is-constructed-of-a-uniform-wire-with-resistance-10-per-cm-the-sides-of-the-letter-are-20-cm-and-the-cross-piece-in-the-middle-is-10-cm-long-96953


tags: ['Physics', 'Dual Nature of Matter', 'BITSAT', 'BITSAT 2012']
source_db: MarksBatch1_P1.db
title:
a-light-having-wavelength-300-nm-fall-on-a-metal-surface-the-work-function-of-metal-is-254-ev-what-is-stopping-potential

slug:
a-light-having-wavelength-300-nm-fall-on-a-metal-surface-the-work-function-of-metal-is-254-ev-what-is-stopping-potential-38483

question:
<div class="question">A light having wavelength $300 \mathrm{nm}$ fall on a metal surface. The work function of metal is $2.54 \mathrm{eV}$, what is stopping potential?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2.3 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$2.59 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$1.59 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$1.29 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$1.59 \mathrm{~V}$</span> </div>

answer:
<div class="solution">Energy of incident radiation is given by:
<br/>\(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.64 * 10^{-34} * 3 * 10^{8}}{300 * 10^{-9}}=6.64 * 10^{-19}=4.05 \mathrm{eV}\)
<br/>The work function of the metal is equal to \(2.54 \mathrm{eV}\)
<br/>Hence the maximum kinetic energy of an ejected photoelectron would be \(1.59 \mathrm{eV}\)
<br/>And the stopping potential needed to stop this would be \(1.59 \mathrm{~V}\).</div>


tags: ['Physics', 'Work Power Energy', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-m--1

slug:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-m-1-32037

question:
<div class="question">A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2},$ find the work done $(\mathrm{in}$
<br/> <br/>joules) by the string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/i7gKSnXmc1MN1Q3bNmoSV1vXdMYrIOUkiYHPtD9sntw.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$4 \mathrm{~J}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$2 \mathrm{~J}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$8 \mathrm{~J}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$10 \mathrm{~J}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$8 \mathrm{~J}$</span> </div>

answer:
<div class="solution">Given $m=0.36 \mathrm{~kg}, M=0.72 \mathrm{~kg}$. The figure shows the forces on $m$ and $M$. When the system is released, let the acceleration be $a$. Then
<br/> <br/>$T-m g=m a$
<br/> <br/>$M g-T=M a$
<br/> <br/>$$
<br/> <br/>a=\frac{(M-m) g}{M+m}=g / 3
<br/> <br/>$$
<br/> <br/>and $T=4 \mathrm{mg} / 3$
<br/> <br/>For block m:
<br/> <br/>$$
<br/> <br/>u=0, a=g / 3, t=1, s=?
<br/> <br/>$$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/_LD6QYhzhQSmiBzK3EKnfY0Rr12eqtvfzy_3HP0QUNg.original.fullsize.png"/><br/> <br/> <br/>$$
<br/> <br/>s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times \frac{g}{3} \times 1^{2}=g / 6
<br/> <br/>$$
<br/> <br/>Work done by the string on $m$ is
<br/> <br/>$$
<br/> <br/>\bar{T} \bar{s}=T s=4 \frac{m g}{3} \times \frac{g}{6}=\frac{4 \times 0.36 \times 10 \times 10}{3 \times 6}=8 \mathrm{~J}
<br/> <br/>$$</div>


tags: ['Physics', 'Work Power Energy', 'BITSAT', 'BITSAT 2013']
source_db: MarksBatch1_P1.db
title:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-m-

slug:
a-light-inextensible-string-that-goes-over-a-smooth-fixed-pulley-as-shown-in-the-figure-connects-two-blocks-of-masses-036-kg-and-072-kg-taking-g-10-m-66074


tags: ['Mathematics', 'Three Dimensional Geometry', 'BITSAT', 'BITSAT 2021']
source_db: MarksBatch1_P1.db
title:
a-line-makes-angles-of-4-5-and-6-0-with-the-positive-axes-of-x-and-y-respectively-the-angle-made-by-the-same-line-with-the-positive-axis-of-z-is

slug:
a-line-makes-angles-of-4-5-and-6-0-with-the-positive-axes-of-x-and-y-respectively-the-angle-made-by-the-same-line-with-the-positive-axis-of-z-is-80699

question:
<div class="question">A line makes angles of $45^{\circ}$ and $60^{\circ}$ with the positive axes of $X$ and $Y$ respectively. The angle made by the same line with the positive axis of $Z$, is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$30^{\circ}$ or $60^{\circ}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$60^{\circ}$ or $90^{\circ}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$90^{\circ}$ or $120^{\circ}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$60^{\circ}$ or $120^{\circ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$60^{\circ}$ or $120^{\circ}$</span> </div>

answer:
<div class="solution">Given $\alpha=45^{\circ}, \beta=60^{\circ}, \gamma=?$
<br/> <br/>$\begin{array}{l}
<br/> <br/>\because \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\
<br/> <br/>\therefore \cos ^{2} \gamma=1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\
<br/> <br/>\Rightarrow \mathrm{y}=60^{\circ} \text { or } 120^{\circ}
<br/> <br/>\end{array}$</div>


tags: ['Mathematics', 'Three Dimensional Geometry', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-line-makes-the-same-angle-with-each-of-the-x-and-z-axes-if-the-angle-which-it-makes-with-yaxis-is-such-that-sin-2-3-sin-2-then-cos-2-equals

slug:
a-line-makes-the-same-angle-with-each-of-the-x-and-z-axes-if-the-angle-which-it-makes-with-yaxis-is-such-that-sin-2-3-sin-2-then-cos-2-equals-32801

question:
<div class="question">A line makes the same angle $\theta$ with each of the $\mathrm{X}$ and $\mathrm{Z}$ -axes. If the angle $\beta$, which it makes with Y-axis, is such that $\sin ^{2} \beta=3 \sin ^{2} \theta,$ then $\cos ^{2} \theta$ equals</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2 / 5$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$1 / 5$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$3 / 5$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$2 / 3$</span> </li> </ul>

answer:
<div class="solution">Let $l, \mathrm{~m}$ and $\mathrm{n}$ be the direction cosines. Then, $l=\cos \theta, \mathrm{m}=\cos \beta, \mathrm{n}=\cos \theta$
<br/> <br/>$$
<br/> <br/>\begin{array}{l}
<br/> <br/>\text { we have } l^{2}+m^{2}+n^{2}=1 \\
<br/> <br/>\Rightarrow \cos ^{2} \theta+\cos ^{2} \beta+\cos ^{2} \theta=1 \\
<br/> <br/>\Rightarrow 2 \cos ^{2} \theta+1-\sin ^{2} \beta=1 \\
<br/> <br/>\Rightarrow 2 \cos ^{2} \theta-\sin ^{2} \beta=0 \\
<br/> <br/>\Rightarrow 2 \cos ^{2} \theta-3 \sin ^{2} \beta=0 \\
<br/> <br/>\quad\left[\because \sin ^{2} \beta=3 \sin ^{2} 0(\text { given })\right] \\
<br/> <br/>\Rightarrow \tan ^{2} \theta=2 / 3 \\
<br/> <br/>\therefore \cos ^{2} \theta=\frac{1}{1+\tan ^{2} \theta}=\frac{1}{1+2 / 3}=\frac{3}{5}
<br/> <br/>\end{array}
<br/> <br/>$$</div>


tags: ['Mathematics', 'Probability', 'BITSAT', 'BITSAT 2009']
source_db: MarksBatch1_P1.db
title:
a-line-segment-has-length-63-and-direction-ratios-are-326-if-the-line-makes-an-obtuse-angle-with-mathrmxaxis-the-components-of-the-line-vector-are

slug:
a-line-segment-has-length-63-and-direction-ratios-are-326-if-the-line-makes-an-obtuse-angle-with-mathrmxaxis-the-components-of-the-line-vector-are-24311

question:
<div class="question">A line segment has length 63 and direction ratios are \(3,-2,6\). If the line makes an obtuse angle with \(\mathrm{x}\)-axis, the components of the line vector are</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">\(27,-18,54\)</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\(-27,18,54\)</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">\(-27,18,-54\)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">\(27,-18,-54\)</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">\(-27,18,-54\)</span> </div>

answer:
<div class="solution">Let the components of the line vector be a, b, c. Then \(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=(63)^2\) ...(i)<br/>Also \(\frac{\mathrm{a}}{3}=\frac{\mathrm{b}}{-2}=\frac{\mathrm{c}}{6}=\lambda\) (say), then \(\mathrm{a}=3 \lambda\), \(\mathrm{b}=-2 \lambda\) and \(\mathrm{c}=6 \lambda\) and from (i) we have \(9 \lambda^2+4 \lambda^2+36 \lambda^2=(63)^2\)<br/>\(\Rightarrow 49 \lambda^2=(63)^2\)<br/>\(\Rightarrow \lambda= \pm \frac{63}{7}= \pm 9\)<br/>Since \(a=3 \lambda  &lt; 0\) as the line makes an obtuse angle with \(x\)-axis, \(\lambda=-9\) and the required components are \(-27,18,-54\).</div>


tags: ['Physics', 'Mechanical Properties of Fluids', 'BITSAT', 'BITSAT 2009']
source_db: MarksBatch1_P1.db
title:
a-liquid-is-flowing-through-a-nonsectional-tube-with-its-axis-horizontally-if-two-points-x-and-y-on-the-axis-of-tube-has-a-sectional-area-20-cm-3-and-

slug:
a-liquid-is-flowing-through-a-nonsectional-tube-with-its-axis-horizontally-if-two-points-x-and-y-on-the-axis-of-tube-has-a-sectional-area-20-cm-3-and-57694

question:
<div class="question">A liquid is flowing through a non-sectional tube with its axis horizontally. If two points $\mathrm{X}$ and $\mathrm{Y}$ on the axis of tube has a sectional area $2.0 \mathrm{~cm}^{3}$ and $25 \mathrm{~mm}^{2}$ respectively then find the flow velocity at $Y$ when the flow velocity at $X$ is $10 \mathrm{~m} / \mathrm{s}$.</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$20 \mathrm{~m} / \mathrm{s}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$40 \mathrm{~m} / \mathrm{s}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$80 \mathrm{~m} / \mathrm{s}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$60 \mathrm{~m} / \mathrm{s}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$80 \mathrm{~m} / \mathrm{s}$</span> </div>

answer:
<div class="solution">According to principle of continuity<br/>$$<br/>v_{y}=\frac{v_{x} A_{x}}{A_{y}}=\frac{10(m / s) \times 2\left(c m^{2}\right)}{25 \times 10^{-2}\left(c m^{2}\right)}=80 \mathrm{~m} / \mathrm{s}<br/>$$</div>


tags: ['Physics', 'Oscillations', 'BITSAT', 'BITSAT 2017']
source_db: MarksBatch1_P1.db
title:
a-load-of-mass-m-falls-from-a-height-h-on-the-scale-pan-hung-from-a-spring-as-shown-if-the-spring-constant-is-k-and-the-mass-of-the-scale-pan-is-zero-

slug:
a-load-of-mass-m-falls-from-a-height-h-on-the-scale-pan-hung-from-a-spring-as-shown-if-the-spring-constant-is-k-and-the-mass-of-the-scale-pan-is-zero-34173

question:
<div class="question"><p>A load of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> falls from a height <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>h</mi></math> on the scale pan hung from a spring as shown. If the spring constant is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>k</mi></math> and the mass of the scale pan is zero and the mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> does not bounce relative to the pan, then the amplitude of vibration is </p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/a5eae703-e0ed-4086-92f2-856e6fc9b23a-image.png" style="width: 200px; height: 306px;"/></p></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac></math></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><msqrt><mn>1</mn><mo>+</mo><mfrac><mrow><mn>2</mn><mi>h</mi><mi>k</mi></mrow><mrow><mi>m</mi><mi>g</mi></mrow></mfrac></msqrt></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><mo>+</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><msqrt><mfrac><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mi>h</mi><mi>k</mi></mrow><mrow><mi>m</mi><mi>g</mi></mrow></mfrac></msqrt></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">None of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><msqrt><mn>1</mn><mo>+</mo><mfrac><mrow><mn>2</mn><mi>h</mi><mi>k</mi></mrow><mrow><mi>m</mi><mi>g</mi></mrow></mfrac></msqrt></math></span> </div>

answer:
<div class="solution"><p>From the energy conservation principle,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>g</mi><mi>h</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>k</mi><msubsup><mi>X</mi><mn>0</mn><mn>2</mn></msubsup><mo>-</mo><mi>m</mi><mi>g</mi><msub><mi>X</mi><mn>0</mn></msub></math></p><p>where, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>X</mi><mn>0</mn></msub></math> is maximum elongation in spring.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>k</mi><msubsup><mi>X</mi><mn>0</mn><mn>2</mn></msubsup><mo>-</mo><mi>m</mi><mi>g</mi><msub><mi>X</mi><mn>0</mn></msub><mo>-</mo><mi>m</mi><mi>g</mi><mi>h</mi><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><msubsup><mi>X</mi><mn>0</mn><mn>2</mn></msubsup><mo>-</mo><mfrac><mrow><mn>2</mn><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><msub><mi>X</mi><mn>0</mn></msub><mo>-</mo><mfrac><mrow><mn>2</mn><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><mi>h</mi><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>X</mi><mn>0</mn></msub><mo>=</mo><mfrac><mrow><mn>2</mn><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mn>2</mn></mfrac><mo>±</mo><msqrt><msup><mfenced><mfrac><mrow><mn>2</mn><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac></mfenced><mn>2</mn></msup><mo>+</mo><mn>4</mn><mo>×</mo><mfrac><mrow><mn>2</mn><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><mi>h</mi></msqrt></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mn>2</mn></mfrac><mo>+</mo><msqrt><msup><mfenced><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac></mfenced><mn>2</mn></msup><mo>+</mo><mfrac><mrow><mn>2</mn><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><mi>h</mi></msqrt></math></p><p>Amplitude <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo></math> elongation in spring for lowest extreme position<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><msub><mi>X</mi><mn>0</mn></msub></mfenced></math> <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo></math> elongation in spring for equilibrium position<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><msub><mi>X</mi><mn>1</mn></msub></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msub><mi>X</mi><mn>0</mn></msub><mo>-</mo><msub><mi>X</mi><mn>1</mn></msub><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac><msqrt><mn>1</mn><mo>+</mo><mfrac><mrow><mn>2</mn><mi>h</mi><mi>k</mi></mrow><mrow><mi>m</mi><mi>g</mi></mrow></mfrac></msqrt><mo> </mo></math>,   <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced close="]" open="["><mrow><mo>∵</mo><msub><mi>X</mi><mn>1</mn></msub><mo>=</mo><mfrac><mrow><mi>m</mi><mi>g</mi></mrow><mi>k</mi></mfrac></mrow></mfenced></math></p></div>


tags: ['Physics', 'Laws of Motion', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-load-of-mass-m-falls-from-a-height-h-on-to-the-scale-pan-hung-from-the-spring-as-shown-in-the-figure-if-the-spring-constant-is-k-and-mass-of-the-sca

slug:
a-load-of-mass-m-falls-from-a-height-h-on-to-the-scale-pan-hung-from-the-spring-as-shown-in-the-figure-if-the-spring-constant-is-k-and-mass-of-the-sca-56014

question:
<div class="question">A load of mass m falls from a height h on to the scale pan hung from the spring as shown in the figure. If the spring constant is $\mathrm{k}$ and mass of the scale pan is zero and the mass $\mathrm{m}$ does not bounce relative to the pan, then the amplitude of vibration is
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/MII3jxvI459IXuFrqsqf4QInRZ3xrVDaHO7VyWKbHHU.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{mg} / \mathrm{d}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{\mathrm{mg}}{\mathrm{k}}+\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}-\frac{\mathrm{mg}}{\mathrm{k}}\right)}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(\frac{1+2 \mathrm{hk}}{\mathrm{mg}}\right)}$</span> </div>

answer:
<div class="solution">According to energy conservation principle, If, $\mathrm{x}_{1}$ is maximum elongation in the spring when the particle is in its lowest extreme position. Then,
<br/> <br/>$$
<br/> <br/>\begin{array}{l}
<br/> <br/>\mathrm{mgh}=\frac{1}{2} \mathrm{kx}_{1}^{2}-\mathrm{mgx}_{1} \\
<br/> <br/>\Rightarrow \frac{1}{2} \mathrm{kx}_{1}^{2}-\mathrm{mgx}_{1}-\mathrm{mgh}=0 \\
<br/> <br/>\text { or, } \mathrm{x}_{1}^{2}-\frac{2 \mathrm{mg}}{\mathrm{k}} \mathrm{x}_{1}-\frac{2 \mathrm{mg}}{\mathrm{k}} \cdot \mathrm{h}=0 \\
<br/> <br/>\therefore \mathrm{x}_{1}=\frac{\frac{2 \mathrm{mg}}{\mathrm{k}} \pm \sqrt{\left.\left(\frac{2 \mathrm{mg}}{\mathrm{k}}\right)^{2}+4 \times \frac{2 \mathrm{mg}}{\mathrm{k}} \mathrm{h}\right]}}{2}
<br/> <br/>\end{array}
<br/> <br/>$$
<br/> <br/>Amplitude $\mathrm{A}=\mathrm{X}_{1}-\mathrm{X}_{0}$ (elongation in spring for equilibrium position)
<br/> <br/>$$
<br/> <br/>\mathrm{A}=\frac{\mathrm{mg}}{\mathrm{k}} \sqrt{\left(1+\frac{2 \mathrm{hk}}{\mathrm{mg}}\right)}
<br/> <br/>$$</div>


tags: ['Physics', 'Magnetic Effects of Current', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-long-straight-wire-along-the-z-axis-carries-a-current-i-in-the-negative-z-direction-the-magnetic-vector-field-b-at-a-point-having-coordinates-x-y-in

slug:
a-long-straight-wire-along-the-z-axis-carries-a-current-i-in-the-negative-z-direction-the-magnetic-vector-field-b-at-a-point-having-coordinates-x-y-in-99640

question:
<div class="question">A long straight wire along the $Z$ -axis carries a current I in the negative $Z$ -direction. The
<br/> <br/>magnetic vector field $\vec{B}$ at a point having coordinates $(x, y)$ in the $Z=0$ plane is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\frac{\mu_{0} I(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{\mu_{0} I(x \hat{i}+y \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{\mu_{0} I(x \hat{j}-y \hat{i})}{2 \pi\left(x^{2}+y^{2}\right)}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{\mu_{0} I(x \hat{i}-y \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{\mu_{0} I(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$</span> </div>

answer:
<div class="solution">The wire carries a current $I$ in the negative $z$ -direction. We have to consider the
<br/> <br/>magnetic vector field $\vec{B}$ at $(x, y)$ in the $z=0$ plane.
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/USmKK_7_XG-H5CPMkrNdWUbIbHIBnjlAwdJk7tBK1OY.original.fullsize.png"/><br/> <br/> <br/>Magnetic field $\vec{B}$ is perpendicular to $O P$ $\therefore \vec{B}=B \sin \theta \hat{i}-B \cos \theta \hat{j}$
<br/> <br/>$$
<br/> <br/>\begin{array}{l}
<br/> <br/>\sin \theta=\frac{y}{r}, \cos \theta=\frac{x}{r} B=\frac{\mu_{0} I}{2 \pi r} \\
<br/> <br/>\therefore \quad \vec{B}=\frac{\mu_{0} I}{2 \pi r^{2}}(y \hat{i}-x \hat{j})
<br/> <br/>\end{array}
<br/> <br/>$$
<br/> <br/>or $\quad \vec{B}=\frac{\mu_{0} I(y \hat{i}-x \hat{j})}{2 \pi\left(x^{2}+y^{2}\right)}$</div>


tags: ['Physics', 'Work Power Energy', 'BITSAT', 'BITSAT 2010']
source_db: MarksBatch1_P1.db
title:
a-long-string-is-stretched-by-2-cm-and-the-potential-energy-is-v-if-the-spring-is-stretched-by-10-cm-its-potential-energy-will-be

slug:
a-long-string-is-stretched-by-2-cm-and-the-potential-energy-is-v-if-the-spring-is-stretched-by-10-cm-its-potential-energy-will-be-78160

question:
<div class="question">A long string is stretched by $2 \mathrm{~cm}$ and the potential energy is $\mathrm{V}$. If the spring is stretched by $10 \mathrm{~cm}$, its potential energy will be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{V} / 25$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{V} / 5$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$5 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$25 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$25 \mathrm{~V}$</span> </div>

answer:
<div class="solution">$\quad V=\frac{1}{2} k(x)^{2}=\frac{1}{2} k(2)^{2}$ or $k=\frac{2 V}{4}=\frac{V}{2}$<br/>$$<br/>\mathrm{V}^{\prime}=\frac{1}{2} \mathrm{k}(10)^{2}=\frac{1}{2} \times\left(\frac{\mathrm{V}}{2}\right)(10)^{2}=25 \mathrm{~V}<br/>$$</div>


tags: ['Physics', 'Work Power Energy', 'BITSAT', 'BITSAT 2022']
source_db: MarksBatch1_P1.db
title:
a-machine-gun-fires-300-bullets-per-min-if-the-mass-of-each-bullet-is-10-g-and-the-velocity-of-the-bullets-is-600-ms-1-the-power-in-kw-of-the-gun-is

slug:
a-machine-gun-fires-300-bullets-per-min-if-the-mass-of-each-bullet-is-10-g-and-the-velocity-of-the-bullets-is-600-ms-1-the-power-in-kw-of-the-gun-is-45750

question:
<div class="question">A machine gun fires 300 bullets per $\mathrm{min}$ if the mass of each bullet is $10 \mathrm{~g}$ and the velocity of the bullets is $600 \mathrm{~ms}^{-1}$, the power (in $\mathrm{kW}$ ) of the gun is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"> <br/>$43200$
<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"> <br/>$9$
<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"> <br/>$72$
<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"> <br/>$7.2$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"> <br/>$9$
<br/></span> </div>

answer:
<div class="solution">Work done by the gun $=$ total kinetic energy of the bullets
<br/>$$
<br/>\begin{aligned}
<br/>&amp; =n \frac{1}{2} m v^2=300 \times \frac{1}{2} \times 10 \times 10^{-3} \times(600)^2 \\
<br/>&amp; =150 \times 10 \times 10^{-3} \times 600 \times 600
<br/>\end{aligned}
<br/>$$
<br/>$$
<br/>\begin{aligned}
<br/>\text { Power of gun } &amp; =\frac{\text { work done }}{\text { Time taken }} \\
<br/>&amp; =\frac{150 \times 10 \times 10^{-3} \times 600 \times 600}{60 \mathrm{~s}} \\
<br/>&amp; =9 \mathrm{~kW} \quad(\because 1 \text { minute }=60 \text { seconds })
<br/>\end{aligned}
<br/>$$</div>


tags: ['Physics', 'Center of Mass Momentum and Collision', 'BITSAT', 'BITSAT 2012']
source_db: MarksBatch1_P1.db
title:
a-man-of-mass-100-kg-is-standing-on-a-platform-of-mass-200-kg-which-is-kept-on-a-smooth-ice-surface-if-the-man-starts-moving-on-the-platform-with-a-sp

slug:
a-man-of-mass-100-kg-is-standing-on-a-platform-of-mass-200-kg-which-is-kept-on-a-smooth-ice-surface-if-the-man-starts-moving-on-the-platform-with-a-sp-14375

question:
<div class="question">A man of mass $100 \mathrm{~kg}$. is standing on a platform of mass $200 \mathrm{~kg}$. which is kept on a smooth ice surface. If the man starts moving on the platform with a speed $30 \mathrm{~m} / \mathrm{sec}$ relative to the platform then calculate with what velocity relative to the ice the platform will recoil?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$5 \mathrm{~m} / \mathrm{sec}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$10 \mathrm{~m} / \mathrm{sec}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$15 \mathrm{~m} / \mathrm{sec}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$20 \mathrm{~m} / \mathrm{sec}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$10 \mathrm{~m} / \mathrm{sec}$</span> </div>

answer:
<div class="solution">Consider the situation shown in figure.Suppose the man moves at a speed w towards right and the the platform recoils at a speed \(\mathrm{V}\) towards left, both relative to the ice. Hence, the speed of the man relative to the platform is
<br/>\(\mathrm{V}+\mathrm{w}\). By the question,
<br/>\(\mathrm{V}+\mathrm{w}=\mathrm{v}\), or \(\mathrm{w}=\mathrm{v}-\mathrm{V} \ldots\) (i)
<br/>Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially both the man and the platform were at rest. Thus,
<br/>\(0=\mathrm{MV}-\underset{m v}{\operatorname{mor}} \mathrm{MV}=\mathrm{m}(\mathrm{v}-\mathrm{V})[\mathrm{Using}(\mathrm{i})]\)
<br/>or, \(V=\frac{m v}{M+m}\)
<br/>Hence, the velocity relative to the ice is \(V=\frac{m v}{M+m}\).</div>


tags: ['Physics', 'Gravitation', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-man-of-mass-m-starts-falling-towards-a-planet-of-mass-m-and-radius-r-as-he-reaches-near-to-the-surface-he-realizes-that-he-will-pass-through-a-small

slug:
a-man-of-mass-m-starts-falling-towards-a-planet-of-mass-m-and-radius-r-as-he-reaches-near-to-the-surface-he-realizes-that-he-will-pass-through-a-small-92224

question:
<div class="question">A man of mass m starts falling towards a planet of mass $\mathrm{M}$ and radius $\mathrm{R}$. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass $2 \mathrm{M} / 3$ and a point mass $\mathrm{M} / 3$ at the centre. Change in the force of gravity experienced by the man is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">0</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{1}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{4}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$</span> </div>

answer:
<div class="solution">$\quad$ Change in force of gravity<br/>$$<br/>=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}-\frac{\mathrm{G} \frac{\mathrm{M}}{3} \mathrm{~m}}{\mathrm{R}^{2}}<br/>$$<br/>(only due to mass $\mathrm{M} / 3$ due to shell gravitational field is zero (inside the shell))<br/>$$<br/>=\frac{2 \mathrm{GMm}}{3 \mathrm{R}^{2}}<br/>$$</div>


tags: ['Physics', 'Motion In One Dimension', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-man-runs-at-a-speed-of-4-m-s-to-overtake-a-standing-bus-when-he-is-6-m-behind-the-door-at-t-0-the-bus-moves-forward-and-continuous-with-a-constant-a-1

slug:
a-man-runs-at-a-speed-of-4-m-s-to-overtake-a-standing-bus-when-he-is-6-m-behind-the-door-at-t-0-the-bus-moves-forward-and-continuous-with-a-constant-a-1-63486

question:
<div class="question">A man runs at a speed of $4 \mathrm{~m} / \mathrm{s}$ to overtake a standing bus. When he is $6 \mathrm{~m}$ behind the door at $\mathrm{t}=0,$ the bus moves forward and continuous with a constant acceleration of $1.2 \mathrm{~m} / \mathrm{s}^{2}$. The man reaches the door in time $\mathrm{t}$. Then,</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$1.2 \mathrm{t}^{2}=4 \mathrm{t}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$4 t^{2}=1.2 t$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$6+4 \mathrm{t}=0.2 \mathrm{t}^{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$</span> </div>

answer:
<div class="solution">Let us draw the figure for given situation,
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/BWqLadMkE5KfR_ErIKoeVL-josDUGvLdgPXyZx0GQ_I.original.fullsize.png"/><br/> <br/> <br/>$\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrm{t}^{2}$
<br/> <br/>$\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$</div>


tags: ['Physics', 'Motion In One Dimension', 'BITSAT', 'BITSAT 2013']
source_db: MarksBatch1_P1.db
title:
a-man-runs-at-a-speed-of-4-m-s-to-overtake-a-standing-bus-when-he-is-6-m-behind-the-door-at-t-0-the-bus-moves-forward-and-continuous-with-a-constant-a

slug:
a-man-runs-at-a-speed-of-4-m-s-to-overtake-a-standing-bus-when-he-is-6-m-behind-the-door-at-t-0-the-bus-moves-forward-and-continuous-with-a-constant-a-57855

question:
<div class="question">A man runs at a speed of $4 \mathrm{~m} / \mathrm{s}$ to overtake a standing bus. When he is $6 \mathrm{~m}$ behind the door at $\mathrm{t}=0,$ the bus moves forward and continuous with a constant acceleration of $1.2 \mathrm{~m} / \mathrm{s}^{2}$. The man reaches the door in time $\mathrm{t}$. Then,</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$1.2 \mathrm{t}^{2}=4 \mathrm{t}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$4 t^{2}=1.2 t$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$6+4 \mathrm{t}=0.2 \mathrm{t}^{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$</span> </div>

answer:
<div class="solution">Let us draw the figure for given situation,
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/BWqLadMkE5KfR_ErIKoeVL-josDUGvLdgPXyZx0GQ_I.original.fullsize.png"/><br/> <br/> <br/>$\Rightarrow \quad 4 \mathrm{t}=6+\frac{1}{2} \times 1.2 \times \mathrm{t}^{2}$
<br/> <br/>$\Rightarrow \quad 4 \mathrm{t}=6+0.6 \mathrm{t}^{2}$</div>


tags: ['Mathematics', 'Probability', 'BITSAT', 'BITSAT 2016']
source_db: MarksBatch1_P1.db
title:
a-man-takes-a-step-forward-with-probability-04-and-backward-with-probability-06-the-probability-that-at-the-end-of-eleven-steps-he-is-one-step-away-fr

slug:
a-man-takes-a-step-forward-with-probability-04-and-backward-with-probability-06-the-probability-that-at-the-end-of-eleven-steps-he-is-one-step-away-fr-98286

question:
<div class="question">A man takes a step forward with probability 0.4 and backward with probability 0.6 . The probability that at the end of eleven steps he is one step away from the starting point is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{2^{5} \cdot 3^{5}}{5^{10}}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$462 \times\left(\frac{6}{25}\right)^{5}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$231 \times \frac{3^{5}}{5^{10}}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">none of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$462 \times\left(\frac{6}{25}\right)^{5}$</span> </div>

answer:
<div class="solution">As $0.4+0.6=1$, the man either takes a step forward or a step backward. Let a step forward be a success and a step backward be a failure. Then, the probability of success in one step
<br/> <br/>$=p=0.4=\frac{2}{5}$
<br/> <br/>The probability of failure in one step
<br/> <br/>$=\mathrm{q}=0.6=\frac{3}{5}$
<br/> <br/>In 11 steps he will be one step away from the starting point if the numbers of successes and failures differ by 1 . So, the number of successes $=6$
<br/> <br/>The number of failures $=5$
<br/> <br/>or the number of successes $=5$, The number of failures $=6$
<br/> <br/>$\therefore$ the required probability
<br/> <br/>$={ }^{11} \mathrm{C}_{6} \mathrm{p}^{6} \mathrm{q}^{5}+{ }^{11} \mathrm{C}_{5} \mathrm{p}^{5} \mathrm{q}^{6}$
<br/> <br/>$={ }^{11} \mathrm{C}_{6}\left(\frac{2}{5}\right)^{6} \cdot\left(\frac{3}{5}\right)^{5}+{ }^{11} \mathrm{C}_{5}\left(\frac{2}{5}\right)^{5}\left(\frac{3}{5}\right)^{6}$
<br/> <br/>$=\frac{11 !}{6 ! 5 !}\left(\frac{2}{5}\right)^{5} \cdot\left(\frac{3}{5}\right)^{5}\left\{\frac{2}{5}+\frac{3}{5}\right\}$
<br/> <br/>$=\frac{11.10 .9 .8 .7}{120} \cdot \frac{2^{5} \cdot 3^{5}}{5^{10}}=462 \times\left(\frac{6}{25}\right)^{5}$</div>


tags: ['Physics', 'Oscillations', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-mass-m-is-suspended-from-a-spring-of-force-constant-k-and-just-touches-another-identical-spring-fixed-to-the-floor-as-shown-in-the-figure-the-time-p-1

slug:
a-mass-m-is-suspended-from-a-spring-of-force-constant-k-and-just-touches-another-identical-spring-fixed-to-the-floor-as-shown-in-the-figure-the-time-p-1-94892

question:
<div class="question">A mass $\mathrm{m}$ is suspended from a spring of force constant $k$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/Ji9m78a_dSgQ2rP5jhAVEJItYhaIPPG6Pcc7xHphjyw.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k} / 2}}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\pi \sqrt{\frac{\mathrm{m}}{3 \mathrm{k} / 2}}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.</span> </div>

answer:
<div class="solution">When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another halfoscillation in the upward direction. The total time period is:<br/>$$<br/>\mathrm{T}=\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}<br/>$$</div>


tags: ['Physics', 'Oscillations', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-mass-m-is-suspended-from-a-spring-of-force-constant-k-and-just-touches-another-identical-spring-fixed-to-the-floor-as-shown-in-the-figure-the-time-p

slug:
a-mass-m-is-suspended-from-a-spring-of-force-constant-k-and-just-touches-another-identical-spring-fixed-to-the-floor-as-shown-in-the-figure-the-time-p-58111

question:
<div class="question">A mass $\mathrm{m}$ is suspended from a spring of force constant $k$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/Ji9m78a_dSgQ2rP5jhAVEJItYhaIPPG6Pcc7xHphjyw.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k} / 2}}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\pi \sqrt{\frac{\mathrm{m}}{3 \mathrm{k} / 2}}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$.</span> </div>

answer:
<div class="solution">When the spring undergoes displacement in the downward direction it completes one half oscillation while it completes another halfoscillation in the upward direction. The total time period is:<br/>$$<br/>\mathrm{T}=\pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}+\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}<br/>$$</div>


tags: ['Physics', 'Work Power Energy', 'BITSAT', 'BITSAT 2011']
answer: <div class="solution">No Solution available</div>
source_db: MarksBatch1_P1.db
title:
a-massless-platform-is-kept-on-a-light-elastic-spring-as-shown-in-fig-when-a-sand-particle-of-mass-01-kg-is-dropped-on-the-pan-from-a-height-of-024-m-

slug:
a-massless-platform-is-kept-on-a-light-elastic-spring-as-shown-in-fig-when-a-sand-particle-of-mass-01-kg-is-dropped-on-the-pan-from-a-height-of-024-m-57642

question:
<div class="question">A massless platform is kept on a light elastic spring as shown in fig. When a sand particle of mass $0.1 \mathrm{~kg}$ is dropped on the pan from a height of $0.24 \mathrm{~m},$ the particle strikes the pan and spring is compressed by $0.01 \mathrm{~m}$. From what height should the particle be dropped to cause a compression of $0.04 \mathrm{~m}$.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/VOZwHqRlKn2xl2kyM8b4t6lL9gxkxmHNIE5NJo7QlXo.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$3.96 \mathrm{~m}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$0.396 \mathrm{~m}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$4 \mathrm{~m}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$0.4 \mathrm{~m}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$0.396 \mathrm{~m}$</span> </div>


tags: ['Physics', 'Dual Nature of Matter', 'BITSAT', 'BITSAT 2010']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">zero</span> </div>
answer: <div class="solution">No Solution available</div>
source_db: MarksBatch1_P1.db
title:
a-material-particle-with-a-rest-mass-m-0-is-moving-with-a-velocity-of-light-c-then-the-wavelength-of-the-de-broglie-wave-associated-with-it-is

slug:
a-material-particle-with-a-rest-mass-m-0-is-moving-with-a-velocity-of-light-c-then-the-wavelength-of-the-de-broglie-wave-associated-with-it-is-88292

question:
<div class="question">A material particle with a rest mass $\mathrm{m}_{0}$ is moving with a velocity of light $\mathrm{c}$. Then the wavelength of the de Broglie wave associated with it is :</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\left(\mathrm{h} / \mathrm{m}_{0} \mathrm{c}\right)$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">zero</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\infty$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\left(\mathrm{m}_{0} \mathrm{c} / \mathrm{h}\right)$</span> </li> </ul>


tags: ['Physics', 'Laws of Motion', 'BITSAT', 'BITSAT 2019']
source_db: MarksBatch1_P1.db
title:
a-maximum-of-3-n-force-can-be-applied-on-a-block-down-the-incline-that-would-still-not-move-the-block-which-is-placed-on-a-rough-inclined-plane-as-sho

slug:
a-maximum-of-3-n-force-can-be-applied-on-a-block-down-the-incline-that-would-still-not-move-the-block-which-is-placed-on-a-rough-inclined-plane-as-sho-96604

question:
<div class="question"><p>A maximum of <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo> </mo><mi mathvariant="normal">N</mi></math> force can be applied on a block, down the incline, that would still not move the block which is placed on a rough inclined plane as shown in the figure. The maximum external force up the inclined plane that does not move the block is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>12</mn><mo> </mo><mi mathvariant="normal">N</mi></math>. The coefficient of static friction between the block and the plane is <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced close="]" open="["><mrow><mi>Take</mi><mo> </mo><mi>g</mi><mo>=</mo><mn>10</mn><mo> </mo><mi mathvariant="normal">m</mi><mo> </mo><msup><mi mathvariant="normal">s</mi><mrow><mo>-</mo><mn>2</mn></mrow></msup></mrow></mfenced></math></p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/3fdda621-387e-4e66-9b7d-a5659fe711f3-image.png" style="width: 180px; height: 130px;"/></p></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msqrt><mn>3</mn></msqrt><mn>5</mn></mfrac></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msqrt><mn>3</mn></msqrt></mfrac></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>3</mn><msqrt><mn>3</mn></msqrt></mrow><mn>5</mn></mfrac></math></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>5</mn><mrow><mn>3</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>5</mn><mrow><mn>3</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac></math></span> </div>

answer:
<div class="solution"><p>Block does not move upto a maximum applied force of <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo> </mo><mi mathvariant="normal">N</mi><mo> </mo></math>down the inclined plane.</p><p>Equation of motion is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>+</mo><mi>m</mi><mi>g</mi><mo> </mo><mi>sin</mi><mn>30</mn><mo>°</mo><mo>=</mo><mi>F</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>+</mo><mi>m</mi><mi>g</mi><mo> </mo><mi>sin</mi><mn>30</mn><mo>°</mo><mo>=</mo><mi>μ</mi><mo> </mo><mi>m</mi><mi>g</mi><mo> </mo><mi>cos</mi><mn>30</mn><mo>°</mo></math>            ....(i)</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/f5d653c4-fa70-46ad-9f2c-5a15aaa73b5d-image.png" style="width: 300px; height: 181px;"/></p><p>Similarly, block also does not move upto a maximum applied force of <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>12</mn><mo> </mo><mi mathvariant="normal">N</mi></math> up the plane.</p><p>Now, equation of motion is</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>g</mi><mo> </mo><mi>sin</mi><mn>30</mn><mo>°</mo><mo>+</mo><mi>F</mi><mo>=</mo><mn>12</mn></math></p><p>or, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>g</mi><mo> </mo><mi>sin</mi><mn>30</mn><mo>°</mo><mo>+</mo><mi>μ</mi><mo> </mo><mi>m</mi><mi>g</mi><mo> </mo><mi>cos</mi><mn>30</mn><mo>°</mo><mo>=</mo><mn>12</mn></math>           ...(ii)</p><p>Solving Eqs. (i) and (ii), we have</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>g</mi><mo> </mo><mi>sin</mi><mn>30</mn><mo>°</mo><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi><mo> </mo><mi>m</mi><mi>g</mi><mo> </mo><mi>cos</mi><mn>30</mn><mo>°</mo><mo>=</mo><mfrac><mn>15</mn><mn>2</mn></mfrac></math></p><p>Hence, dividing</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mfrac><mn>1</mn><mi>μ</mi></mfrac><mi>tan</mi><mn>30</mn><mo>°</mo><mo>=</mo><mfrac><mrow><mn>9</mn><mo>/</mo><mn>2</mn></mrow><mrow><mn>15</mn><mo>/</mo><mn>2</mn></mrow></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>5</mn></mfrac><mspace linebreak="newline"></mspace></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>μ</mi><mo>=</mo><mfrac><mn>5</mn><mrow><mn>3</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac></math></p></div>


tags: ['Physics', 'Electromagnetic Induction', 'BITSAT', 'BITSAT 2021']
source_db: MarksBatch1_P1.db
title:
a-metal-disc-of-radius-100-cm-is-rotated-at-a-constant-angular-speed-of-60-rad-s-in-a-plane-at-right-angles-to-an-external-field-of-magnetic-induction

slug:
a-metal-disc-of-radius-100-cm-is-rotated-at-a-constant-angular-speed-of-60-rad-s-in-a-plane-at-right-angles-to-an-external-field-of-magnetic-induction-82015

question:
<div class="question">A metal disc of radius $100 \mathrm{~cm}$ is rotated at a constant angular speed of $60 \mathrm{rad} / \mathrm{s}$ in a plane at right angles to an external field of magnetic induction $0.05 \mathrm{~Wb} / \mathrm{m}^{2}$. The emf induced between the centre and a point on the rim will be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$3 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$1.5 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$6 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$9 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$1.5 \mathrm{~V}$</span> </div>

answer:
<div class="solution">Induced emf produced between the centre and a point on the disc is given by
<br/> <br/>$\begin{array}{l}
<br/> <br/>e=\frac{1}{2} \omega \mathrm{BR}^{2} \\
<br/> <br/>\omega=60 \mathrm{rad} / \mathrm{s}, \mathrm{B}=0.05 \mathrm{~Wb} / \mathrm{m}^{2} \\
<br/> <br/>\text { and } \mathrm{R}=100 \mathrm{~cm}=1 \mathrm{~m} \\
<br/> <br/>\text { We get } e=\frac{1}{2} \times 60 \times 0.05 \times(1)^{2}=1.5 \mathrm{~V}
<br/> <br/>\end{array}$</div>


tags: ['Physics', 'Electromagnetic Induction', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-metal-rod-of-length-1-m-is-rotated-about-one-of-its-ends-in-a-plane-right-angles-to-a-field-of-inductance-25-1-0-3-wb-m-2-if-it-makes-1800-revolutio

slug:
a-metal-rod-of-length-1-m-is-rotated-about-one-of-its-ends-in-a-plane-right-angles-to-a-field-of-inductance-25-1-0-3-wb-m-2-if-it-makes-1800-revolutio-89196

question:
<div class="question">A metal rod of length $1 \mathrm{~m}$ is rotated about one of its ends in a plane right angles to a field of inductance $2.5 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}$. If it makes 1800 revolutions/min. Calculate induced e.m.f. between its ends.</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2.471 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$3.171 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$0.471 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$1.771 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$0.471 \mathrm{~V}$</span> </div>

answer:
<div class="solution">Given : $\ell=1 \mathrm{~m}, \mathrm{~B}=5 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}$<br/>$$<br/>\mathrm{f}=\frac{1800}{60}=30 \text { rotations } / \mathrm{sec}<br/>$$<br/>In one rotation, the moving rod of the metal traces a circle of radius $\mathrm{r}=\ell$<br/>$\therefore$ Area swept in one rotation $=\pi r^{2}$<br/>$$<br/>\begin{array}{l}<br/>\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})=\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{B} \pi \mathrm{r}^{2}}{\mathrm{~T}} \\<br/>=\mathrm{B} \mathrm{f} \pi \mathrm{r}^{2}=\left(5 \times 10^{-3}\right) \times 3.14 \times 30 \times 1=0.471 \mathrm{~V}<br/>\end{array}<br/>$$</div>


tags: ['Chemistry', 'Solid State', 'BITSAT', 'BITSAT 2018']
source_db: MarksBatch1_P1.db
title:
a-metallic-element-has-a-cubic-lattice-each-edge-of-the-unit-cell-is-2-a-and-the-density-of-metal-is-25-g-cm-3-the-unit-cell-in-200-g-of-metal-are

slug:
a-metallic-element-has-a-cubic-lattice-each-edge-of-the-unit-cell-is-2-a-and-the-density-of-metal-is-25-g-cm-3-the-unit-cell-in-200-g-of-metal-are-37113

question:
<div class="question">A metallic element has a cubic lattice. Each edge of the unit cell is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo> </mo><mover><mi mathvariant="normal">A</mi><mo>.</mo></mover></math> and the density of metal is <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>25</mn><mo> </mo><mi mathvariant="normal">g</mi><mo> </mo><msup><mi>cm</mi><mn>3</mn></msup></math>. The unit cell in <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>200</mn><mo> </mo><mi mathvariant="normal">g</mi></math> of metal are</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>×</mo><msup><mn>10</mn><mn>24</mn></msup></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>×</mo><msup><mn>10</mn><mn>22</mn></msup></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>×</mo><msup><mn>10</mn><mn>20</mn></msup></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>×</mo><msup><mn>10</mn><mn>25</mn></msup></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>×</mo><msup><mn>10</mn><mn>24</mn></msup></math></span> </div>

answer:
<div class="solution"><p>Edge length of the unit cell <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>2</mn><mover><mi mathvariant="normal">A</mi><mo>.</mo></mover><mo>=</mo><mn>2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>8</mn></mrow></msup><mo> </mo><mi>cm</mi></math></p><p>Volume of unit cell <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msup><mfenced><mrow><mn>2</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>8</mn></mrow></msup></mrow></mfenced><mn>3</mn></msup><mo>=</mo><mn>8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>24</mn></mrow></msup><mo> </mo><msup><mi>cm</mi><mn>3</mn></msup></math></p><p>Mass of one unit cell <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo></math> volume <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>×</mo></math> density</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>24</mn></mrow></msup><mo>×</mo><mn>25</mn></math></p><p>Number of unit cell in <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>200</mn><mo> </mo><mi mathvariant="normal">g</mi></math> of metal</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mtext> mass of the metal </mtext><mtext> mass of one unit cell </mtext></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mn>200</mn><mrow><mn>8</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>24</mn></mrow></msup><mo>×</mo><mn>25</mn></mrow></mfrac><mo>=</mo><mn>1</mn><mo>×</mo><msup><mn>10</mn><mn>24</mn></msup></math></p></div>


tags: ['Physics', 'Mechanical Properties of Solids', 'BITSAT', 'BITSAT 2009']
source_db: MarksBatch1_P1.db
title:
a-metallic-rod-breaks-when-strain-produced-is-02-the-youngs-modulus-of-the-material-of-the-rod-is-7-1-0-9-n-m-2-what-should-be-its-area-of-crosssectio

slug:
a-metallic-rod-breaks-when-strain-produced-is-02-the-youngs-modulus-of-the-material-of-the-rod-is-7-1-0-9-n-m-2-what-should-be-its-area-of-crosssectio-27171

question:
<div class="question">A metallic rod breaks when strain produced is $0.2 \% .$ The Young's modulus of the material of the rod is $7 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} .$ What should be its area of cross-section to support a load of $10^{4} \mathrm{~N}$ ?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$7.1 \times 10^{-8} \mathrm{~m}^{2}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$7.1 \times 10^{-6} \mathrm{~m}^{2}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$7.1 \times 10^{-4} \mathrm{~m}^{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$7.1 \times 10^{-2} \mathrm{~m}^{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$7.1 \times 10^{-4} \mathrm{~m}^{2}$</span> </div>

answer:
<div class="solution">$\quad$ Maximum possible strain $=0.2 / 100$<br/>$$<br/>\begin{array}{l}<br/>\therefore \mathrm{A}=\frac{\mathrm{F}}{\mathrm{Y} \times \text { strain }} \\<br/>=\frac{10^{4} \times 100}{\left(7 \times 10^{9}\right) \times 0.2}=7.1 \times 10^{-4} \mathrm{~m}^{2}<br/>\end{array}<br/>$$</div>


tags: ['Physics', 'Wave Optics', 'BITSAT', 'BITSAT 2012']
source_db: MarksBatch1_P1.db
title:
a-mica-slit-of-thickness-t-and-refractive-index-is-introduoed-in-theray-from-the-first-source-s-1-by-how-much-distance-of-fringes-pattern-will-be-disp

slug:
a-mica-slit-of-thickness-t-and-refractive-index-is-introduoed-in-theray-from-the-first-source-s-1-by-how-much-distance-of-fringes-pattern-will-be-disp-63165

question:
<div class="question">A mica slit of thickness t and refractive index $\mu$ is introduoed in theray from the first source S$_{1}$. By how much distance of fringes pattern will be displaced?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{\mathrm{d}}{\mathrm{D}}(\mu-1) \mathrm{t}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{\mathrm{d}}{(\mu-1) \mathrm{D}}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{D}{d}(\mu-1)$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}$</span> </div>

answer:
<div class="solution">Hint:
<br/>The laws of Electromagnetic Induction give us idea about the various ways in which we can produce Electric Current using Electric or Magnetic field.
<br/>Correct Option is D.
<br/>Explanation for correct answer:
<br/>- Galvanometer is a device which detects current in the circuit.
<br/>- Voltmeter is a device which measures potential difference between two points in the circuit.
<br/>- Electric motor is a device which converts Electrical Energy into Mechanical Energy.
<br/>- Electric Generator is an device which produces Electric Current from Mechanical energy, Chemical energy, or by using Magnetic or Electric Fields. The laws if Electromagnetic Induction are used in Electric Generator which produces Electric Current from Magnetic or Electric Fields.</div>


tags: ['Physics', 'Mechanical Properties of Solids', 'BITSAT', 'BITSAT 2017']
source_db: MarksBatch1_P1.db
title:
a-mild-steel-wire-of-length-2-l-and-crosssectional-area-a-is-stretched-well-with-in-the-elastic-limit-horizontally-between-two-pillars-as-shown-in-fig

slug:
a-mild-steel-wire-of-length-2-l-and-crosssectional-area-a-is-stretched-well-with-in-the-elastic-limit-horizontally-between-two-pillars-as-shown-in-fig-36662

question:
<div class="question"><p>A mild steel wire of length <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo> </mo><mi>L</mi></math> and cross-sectional area <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi></math> is stretched, well with in the elastic limit, horizontally between two pillars as shown in figure. A mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> is suspended from the mid-point of the wire strain in the wire is</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/af66ac02-9463-4916-9de2-0a990960e650-image.png" style="width: 240px; height: 150px;"/></p></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msup><mi>x</mi><mn>2</mn></msup><mrow><mn>2</mn><mo> </mo><msup><mi>L</mi><mn>2</mn></msup></mrow></mfrac></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>x</mi><mi>L</mi></mfrac></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msup><mi>x</mi><mn>2</mn></msup><mi>L</mi></mfrac></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msup><mi>x</mi><mn>2</mn></msup><mrow><mn>2</mn><mo> </mo><mi>L</mi></mrow></mfrac></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msup><mi>x</mi><mn>2</mn></msup><mrow><mn>2</mn><mo> </mo><msup><mi>L</mi><mn>2</mn></msup></mrow></mfrac></math></span> </div>

answer:
<div class="solution"><p>Increase in length </p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/eae578bf-3df1-4616-948e-757cca5971f6-image.png" style="width: 330px; height: 115px;"/></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">Δ</mi><mo> </mo><mi>L</mi><mo>=</mo><mi>B</mi><mi>O</mi><mo>+</mo><mi>O</mi><mi>C</mi><mo>-</mo><mfenced><mrow><mi>B</mi><mi>C</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>2</mn><mi>B</mi><mi>O</mi><mo>-</mo><mn>2</mn><mi>B</mi><mi>D</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi mathvariant="normal">Δ</mi><mi>L</mi><mo>=</mo><mn>2</mn><mi>B</mi><mi>O</mi><mo>-</mo><mn>2</mn><mo> </mo><mi>L</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>2</mn><mo> </mo><msup><mfenced close="]" open="["><mrow><msup><mi>L</mi><mn>2</mn></msup><mo>+</mo><msup><mi>x</mi><mn>2</mn></msup></mrow></mfenced><mfrac><mn>1</mn><mn>2</mn></mfrac></msup><mo>-</mo><mn>2</mn><mo> </mo><mi>L</mi></math></p><p>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">Δ</mi><mi>L</mi><mo>=</mo><mn>2</mn><mo> </mo><mi>L</mi><mo> </mo><msup><mfenced close="]" open="["><mrow><mn>1</mn><mo>+</mo><mfrac><msup><mi>x</mi><mn>2</mn></msup><msup><mi>L</mi><mn>2</mn></msup></mfrac></mrow></mfenced><mfrac><mn>1</mn><mn>2</mn></mfrac></msup><mo>-</mo><mn>2</mn><mo> </mo><mi>L</mi></math> [<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∵</mo></math> using Bionomial theorem]</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi mathvariant="normal">Δ</mi><mi>L</mi><mo>≈</mo><mn>2</mn><mo> </mo><mi>L</mi><mo> </mo><mfenced close="]" open="["><mrow><mn>1</mn><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mfrac><msup><mi>x</mi><mn>2</mn></msup><msup><mi>L</mi><mn>2</mn></msup></mfrac><mo>-</mo><mn>1</mn></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><msup><mi>x</mi><mn>2</mn></msup><mi>L</mi></mfrac></math>   <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced close="]" open="["><mrow><mo>∵</mo><mo> </mo><mi>x</mi><mo>&lt;</mo><mo>&lt;</mo><mi>L</mi></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo></math> Strain <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mrow><mi mathvariant="normal">Δ</mi><mi>L</mi></mrow><mrow><mn>2</mn><mo> </mo><mi>L</mi></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mn>1</mn><mrow><mn>2</mn><mo> </mo><mi>L</mi></mrow></mfrac><mo>×</mo><mo>∆</mo><mi>L</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mn>2</mn><mo> </mo><mi>L</mi></mrow></mfrac><mo>×</mo><mfrac><msup><mi>x</mi><mn>2</mn></msup><mi>L</mi></mfrac><mo>=</mo><mfrac><msup><mi>x</mi><mn>2</mn></msup><mrow><mn>2</mn><mo> </mo><msup><mi>L</mi><mn>2</mn></msup></mrow></mfrac></math>.</p></div>


tags: ['Chemistry', 'p Block Elements (Group 13 & 14)', 'BITSAT', 'BITSAT 2018']
source_db: MarksBatch1_P1.db
title:
a-mixture-of-boron-trichloride-and-hydrogen-is-subjected-to-silent-electric-discharge-to-form-x-and-hcl-x-is-mixed-with-nh-3-and-heated-to-200-0-c-to-

slug:
a-mixture-of-boron-trichloride-and-hydrogen-is-subjected-to-silent-electric-discharge-to-form-x-and-hcl-x-is-mixed-with-nh-3-and-heated-to-200-0-c-to-36808

question:
<div class="question">A mixture of Boron trichloride and hydrogen is subjected to silent electric discharge to form <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">x</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>HCl</mi></math>. <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">x</mi></math> is mixed with <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>NH</mi><mn>3</mn></msub></math> and heated to <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mn>200</mn><mn>0</mn></msup><mi mathvariant="normal">C</mi></math> to form <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">y</mi></math>. The formula of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">y</mi></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">B</mi><mn>2</mn></msub><msub><mi mathvariant="normal">O</mi><mn>3</mn></msub></math></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">B</mi><mn>3</mn></msub><msub><mi mathvariant="normal">N</mi><mn>3</mn></msub><msub><mi mathvariant="normal">H</mi><mn>6</mn></msub></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">H</mi><mn>3</mn></msub><msub><mi>BO</mi><mn>3</mn></msub></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">B</mi><mn>2</mn></msub><msub><mi mathvariant="normal">H</mi><mn>6</mn></msub></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">B</mi><mn>3</mn></msub><msub><mi mathvariant="normal">N</mi><mn>3</mn></msub><msub><mi mathvariant="normal">H</mi><mn>6</mn></msub></math></span> </div>

answer:
<div class="solution"><p>When a mixture of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>BCl</mi><mn>3</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub></math> is subjected to silent electric discharge, diborane and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>HCl</mi></math> are formed. Diborane reacts with <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>NH</mi><mn>3</mn></msub></math> at <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>200</mn><mo>°</mo><mi mathvariant="normal">C</mi></math> to give Borazine (inorganic benzene).</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><msub><mi>BCl</mi><mn>3</mn></msub><mo>+</mo><mn>6</mn><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub><munderover><mo>→</mo><mi>discharge</mi><mi>Electric</mi></munderover><munder><mrow><msub><mi mathvariant="normal">B</mi><mn>2</mn></msub><msub><mi mathvariant="normal">H</mi><mn>6</mn></msub></mrow><mfenced><mi mathvariant="normal">x</mi></mfenced></munder><mo>+</mo><mn>6</mn><mi>HCl</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><munder><mrow><mn>3</mn><msub><mi mathvariant="normal">B</mi><mn>2</mn></msub><msub><mi mathvariant="normal">H</mi><mn>6</mn></msub></mrow><mn>1</mn></munder><mo>+</mo><munder><mrow><mn>6</mn><msub><mi>NH</mi><mn>3</mn></msub></mrow><mn>2</mn></munder><mover><mo>⟶</mo><mrow><mn>200</mn><mo>°</mo><mi mathvariant="normal">C</mi></mrow></mover><munder><mrow><mn>2</mn><msub><mi mathvariant="normal">B</mi><mn>3</mn></msub><msub><mi mathvariant="normal">N</mi><mn>3</mn></msub><msub><mi mathvariant="normal">H</mi><mn>6</mn></msub></mrow><mrow><mi>Borazine</mi><mo>(</mo><mi mathvariant="normal">y</mi><mo>)</mo></mrow></munder><mo>+</mo><mn>12</mn><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub></math></p></div>


tags: ['Chemistry', 'Practical Chemistry', 'BITSAT', 'BITSAT 2013']
source_db: MarksBatch1_P1.db
title:
a-mixture-of-chlorides-of-copper-cadmium-chromium-iron-and-aluminium-was-dissolved-in-water-acidified-with-hcl-and-hydrogen-sulphide-gas-was-passed-fo

slug:
a-mixture-of-chlorides-of-copper-cadmium-chromium-iron-and-aluminium-was-dissolved-in-water-acidified-with-hcl-and-hydrogen-sulphide-gas-was-passed-fo-43808

question:
<div class="question">A mixture of chlorides of copper, cadmium, chromium, iron and aluminium was dissolved in water acidified with $\mathrm{HCl}$ and hydrogen sulphide gas was passed for sufficient time. It was filtered, boiled and a few drops of nitric acid were added while boiling. To this solution ammonium chloride and sodium hydroxide were added and filtered. The filterate shall give test for.</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Sodium and iron</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">Sodium and aluminium</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">Aluminium and iron</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">Sodium, iron, cadmium and Al</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">Sodium and aluminium</span> </div>

answer:
<div class="solution">CuS and CdS are precipitated by $\mathrm{H}_{2} \mathrm{~S}$. Hydroxide of Al will pass into the solution in the form of $\mathrm{NaAlO}_{2}$ being amphoteric in nature. Hence filtrate will give test for sodium and aluminium.</div>


tags: ['Physics', 'Kinetic Theory of Gases', 'BITSAT', 'BITSAT 2022']
source_db: MarksBatch1_P1.db
title:
a-monoatomic-gas-of-molar-mass-m-is-kept-in-a-insulated-container-container-is-moving-with-velocity-v-if-the-container-is-suddenly-stopped-then-the-ch

slug:
a-monoatomic-gas-of-molar-mass-m-is-kept-in-a-insulated-container-container-is-moving-with-velocity-v-if-the-container-is-suddenly-stopped-then-the-ch-78587

question:
<div class="question">A monoatomic gas of molar mass $m$ is kept in a insulated container. Container is moving with velocity $v$. If the container is suddenly stopped, then the change in the temperature of the gas is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"> <br/>$m v^2 / 4 R$
<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"> <br/>$m v^2 / 2 R$
<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"> <br/>$m v^2 / R$
<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"> <br/>$m v^2 / 3 R$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"> <br/>$m v^2 / 3 R$</span> </div>

answer:
<div class="solution">Let the insulated container has $n$ moles of the monoatomic gas. The loss of kinetic energy of the gas,
<br/>$$
<br/>\Delta E_K=\frac{1}{2}(m n) v^2
<br/>$$
<br/>If the change in temperature of the gas is $\Delta T$, then heat gained by the gas is,
<br/>$$
<br/>\Delta Q=\frac{3}{2} n R \Delta T
<br/>$$
<br/>Now, according to question,
<br/>$$
<br/>\begin{aligned}
<br/>\Delta Q &amp; =\Delta E_K \\
<br/>\frac{3}{2} n R \Delta T &amp; =\frac{1}{2} m n v^2 \\
<br/>\Rightarrow \quad \Delta T &amp; =\frac{m v^2}{3 R}
<br/>\end{aligned}
<br/>$$</div>


tags: ['Physics', 'Ray Optics', 'BITSAT', 'BITSAT 2022']
source_db: MarksBatch1_P1.db
title:
a-mosquito-o-is-sitting-infront-of-a-glass-rod-having-spherical-end-of-radius-of-curvature-40-cm-the-image-would-be-formed-at

slug:
a-mosquito-o-is-sitting-infront-of-a-glass-rod-having-spherical-end-of-radius-of-curvature-40-cm-the-image-would-be-formed-at-84535

question:
<div class="question">A mosquito $O$ is sitting infront of a glass rod having spherical end of radius of curvature $40 \mathrm{~cm}$. The image would be formed at
<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/weLGNMYc7iY8HKVyLvwJjxS4kZCrETIc0bLDHgLJXiY.original.fullsize.png"/><br/></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"> <br/>$40 \mathrm{~cm}$ left
<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"> <br/>infinity
<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"> <br/>$20 \mathrm{~cm}$ to the right
<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"> <br/>$15 \mathrm{~cm}$ to the left</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"> <br/>$40 \mathrm{~cm}$ left
<br/></span> </div>

answer:
<div class="solution">As, we know, $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$
<br/>$$
<br/>\begin{aligned}
<br/>&amp; \Rightarrow \quad \frac{1.5}{v}-\frac{1}{(-20)}=\frac{1.5-1}{+40} \\
<br/>&amp; v=-40 \mathrm{~cm} \\
<br/>&amp;
<br/>\end{aligned}
<br/>$$
<br/>Negative sign shows that image is obtained on the same side of object i.e., towards left.</div>


tags: ['Physics', 'Current Electricity', 'BITSAT', 'BITSAT 2021']
source_db: MarksBatch1_P1.db
title:
a-moving-coil-galvanometer-has-resistance-of-10-and-full-scale-deflection-of-001-a-it-can-be-converted-into-voltmeter-of-10-v-full-scale-by-connecting

slug:
a-moving-coil-galvanometer-has-resistance-of-10-and-full-scale-deflection-of-001-a-it-can-be-converted-into-voltmeter-of-10-v-full-scale-by-connecting-91515

question:
<div class="question">A moving coil galvanometer has resistance of $10 \Omega$ and full scale deflection of $0.01 \mathrm{~A}$. It can be converted into voltmeter of $10 \mathrm{~V}$ full scale by connecting into resistance of</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$9.90 \Omega$ in series</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$10 \Omega$ in series</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$990 \Omega$ in series</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$0.10 \Omega$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$990 \Omega$ in series</span> </div>

answer:
<div class="solution">$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}=\frac{10}{0.01}-10=990 \Omega$ in series.</div>


tags: ['Physics', 'Magnetic Effects of Current', 'BITSAT', 'BITSAT 2014']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">1: 1</span> </div>
source_db: MarksBatch1_P1.db
title:
a-narrow-beam-of-protons-and-deuterons-each-having-the-same-momentum-enters-a-region-of-uniform-magnetic-field-directed-perpendicular-to-their-directi

slug:
a-narrow-beam-of-protons-and-deuterons-each-having-the-same-momentum-enters-a-region-of-uniform-magnetic-field-directed-perpendicular-to-their-directi-43871

question:
<div class="question">A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The ratio of the radii of the circular paths described by them is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">1: 2</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">1: 1</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">2: 1</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">1: 3</span> </li> </ul>

answer:
<div class="solution">Since, the radius of circular path of a charged
<br/> <br/>particle in magnetic field is $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\rho}{\mathrm{qB}}$
<br/> <br/>Now, the radius of circular path of charged particle of given momentum $\rho$ and magnetic field $\mathrm{B}$ is given by $\mathrm{r} \propto \frac{1}{\mathrm{q}}$
<br/> <br/>But charge on both charged particles, protons and deuterons, is same. Therefore,
<br/> <br/>$$
<br/> <br/>\frac{\mathrm{r}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{D}}}=\frac{\mathrm{q}_{\mathrm{D}}}{\mathrm{q}_{\rho}}=\frac{1}{\mathrm{l}}
<br/> <br/>$$</div>


tags: ['Physics', 'Center of Mass Momentum and Collision', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-neutron-moving-with-speed-v-makes-a-head-on-collision-with-a-hydrogen-atom-in-ground-state-kept-at-rest-the-minimum-kinetic-energy-of-the-neutron-fo

slug:
a-neutron-moving-with-speed-v-makes-a-head-on-collision-with-a-hydrogen-atom-in-ground-state-kept-at-rest-the-minimum-kinetic-energy-of-the-neutron-fo-81404

question:
<div class="question">A neutron moving with speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$10.2 \mathrm{eV}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$20.4 \mathrm{eV}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$12.1 \mathrm{eV}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$16.8 \mathrm{eV}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$20.4 \mathrm{eV}$</span> </div>

answer:
<div class="solution">Let speed of neutron before collision $=\mathrm{V}$ Speed of neutron after collision $=\mathrm{V}_{1}$ Speed of proton or hydrogen atom after collision $=\mathrm{V}_{2}$ Energy of excitation $=\Delta \mathrm{E}$ From the law of conservation of linear momentum, $\mathrm{mv}=\mathrm{mv}_{1}+\mathrm{mv}_{2}$
<br/> <br/>And for law of conservation of energy,
<br/> <br/>$$
<br/> <br/>\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2} \mathrm{mv}_{2}^{2}
<br/> <br/>$$
<br/> <br/>From squaring eq. (i), we get
<br/> <br/>$$
<br/> <br/>v^{2}=v_{1}^{2}+v_{2}^{2}+2 v_{1} v_{2}
<br/> <br/>$$
<br/> <br/>From squaring eq. (ii), we get
<br/> <br/>$$
<br/> <br/>v^{2}=v_{1}^{2}+v_{2}^{2}+\frac{2 \Delta E}{m}
<br/> <br/>$$
<br/> <br/>From eqn (3) \&amp; (4)
<br/> <br/>$$
<br/> <br/>\begin{aligned}
<br/> <br/>\therefore 2 \mathrm{v}_{1} \mathrm{v}_{2} &amp;=\frac{2 \Delta \mathrm{E}}{\mathrm{m}} \\
<br/> <br/>\therefore\left(\mathrm{v}_{1}-\mathrm{v}_{2}\right)^{2} &amp;=\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)^{2}-4 \mathrm{v}_{1} \mathrm{v}_{2}=\mathrm{v}^{2}-\frac{4 \Delta \mathrm{E}}{\mathrm{m}}
<br/> <br/>\end{aligned}
<br/> <br/>$$
<br/> <br/>As, $v_{1}-v_{2}$ must be real, $v^{2}-\frac{4 \Delta E}{m} \geq 0$
<br/> <br/>$$
<br/> <br/>\Rightarrow \quad \frac{1}{2} \mathrm{mv}^{2} \geq 2 \Delta \mathrm{E}
<br/> <br/>$$
<br/> <br/>The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is $10.2 \mathrm{eV}$. Therefore, the maximum kinetic energy needed is
<br/> <br/>$$
<br/> <br/>\frac{1}{2} m v_{\min }^{2}=2 \times 10.2=20.4 \mathrm{eV}
<br/> <br/>$$</div>


tags: ['Mathematics', 'Parabola', 'BITSAT', 'BITSAT 2022']
source_db: MarksBatch1_P1.db
title:
a-normal-is-drawn-at-the-point-p-to-the-parabola-y-2-8-x-which-is-inclined-at-6-0-with-the-straight-line-y-8-then-the-point-p-lies-on-the-straight-lin

slug:
a-normal-is-drawn-at-the-point-p-to-the-parabola-y-2-8-x-which-is-inclined-at-6-0-with-the-straight-line-y-8-then-the-point-p-lies-on-the-straight-lin-19213

question:
<div class="question">A normal is drawn at the point $P$ to the parabola $y^2=8 x$, which is inclined at $60^{\circ}$ with the straight line $y=8$. Then the point $P$ lies on the straight line</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"> <br/>$2 x+y-12-4 \sqrt{3}=0$
<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"> <br/>$2 x-y-12+4 \sqrt{3}=0$
<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"> <br/>$2 x-y-12-4 \sqrt{3}=0$
<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"> <br/>None of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"> <br/>$2 x-y-12-4 \sqrt{3}=0$
<br/></span> </div>

answer:
<div class="solution">For the parabola $y^2=4 a x$, the equation of normal at $P\left(a m^2,-2 a m\right)$ is $y=m x-2 a m-a m^3$.
<br/>Here, $m=\tan 60^{\circ}=\sqrt{3}$
<br/>$\therefore P \equiv\left(a(\sqrt{3})^2,-2 a(\sqrt{3})\right) \equiv(6,-4 \sqrt{3}) \quad(\because a=2)$
<br/>Thus, $P$ satisfies $2 x-y-12-4 \sqrt{3}=0$</div>


title: a-nucleus-of-uranium-decays-at-rest-into-nuclei-of-thorium-and-helium-then
slug: a-nucleus-of-uranium-decays-at-rest-into-nuclei-of-thorium-and-helium-then-61007
tags: ['Physics', 'Nuclear Physics', 'BITSAT', 'BITSAT 2014']
source_db: MarksBatch1_P1.db
question:
<div class="question">A nucleus of uranium decays at rest into nuclei of thorium and helium. Then :</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">the helium nucleus has less momentum than the thorium nucleus</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">the helium nucleus has more momentum than the thorium nucleus</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">the helium nucleus has less kinetic energy than the thorium nucleus</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">the helium nucleus has more kinetic energy than the thorium nucleus</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">the helium nucleus has more kinetic energy than the thorium nucleus</span> </div>

answer:
<div class="solution">In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity. $\mathrm{U} \rightarrow \mathrm{Th}+\mathrm{He}$
<br/> <br/>$$
<br/> <br/>\mathrm{KE}_{\mathrm{Th}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\mathrm{Th}}}, \mathrm{KE}_{\mathrm{He}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\mathrm{He}}}
<br/> <br/>$$
<br/> <br/>sinc $\mathrm{m}_{\mathrm{He}}$ is less so $\mathrm{KE}_{\mathrm{He}}$ will be more.</div>


tags: ['Physics', 'Center of Mass Momentum and Collision', 'BITSAT', 'BITSAT 2020']
source_db: MarksBatch1_P1.db
title:
a-nucleus-splits-into-two-nuclear-parts-which-have-their-velocity-ratio-equal-to-2-1-what-will-be-the-ratio-of-their-nuclear-radius

slug:
a-nucleus-splits-into-two-nuclear-parts-which-have-their-velocity-ratio-equal-to-2-1-what-will-be-the-ratio-of-their-nuclear-radius-16525

question:
<div class="question">A nucleus splits into two nuclear parts which have their velocity ratio equal to $2: 1 .$ What will be the ratio of their nuclear radius?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2^{1 / 3}: 1$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$1: 2^{1 / 3}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$3^{1 / 2}: 1$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$1: 3^{1 / 2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$1: 2^{1 / 3}$</span> </div>

answer:
<div class="solution">As momentum is conserved, therefore
<br/> <br/>$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\frac{1}{2}$
<br/> <br/>$\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}=\left(\frac{1}{2}\right)^{1 / 3}=1: 2^{1 / 3}$</div>


tags: ['Mathematics', 'Circle', 'BITSAT', 'BITSAT 2013']
source_db: MarksBatch1_P1.db
title:
a-pair-of-tangents-are-drawn-from-the-origin-to-the-circle-x-2-y-2-20-x-y-20-0-then-the-equation-of-the-pair-of-tangent-are

slug:
a-pair-of-tangents-are-drawn-from-the-origin-to-the-circle-x-2-y-2-20-x-y-20-0-then-the-equation-of-the-pair-of-tangent-are-23632

question:
<div class="question">A pair of tangents are drawn from the origin to the circle $x^{2}+y^{2}+20(x+y)+20=0,$ then the equation of the pair of tangent are</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$x^{2}+y^{2}-5 x y=0$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$x^{2}+y^{2}+2 x+y=0$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$x^{2}+y^{2}-x y+7=0$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$2 x^{2}+2 y^{2}+5 x y=0$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$2 x^{2}+2 y^{2}+5 x y=0$</span> </div>

answer:
<div class="solution">Equation of pair of tangents is given by $\mathrm{SS}_{1}=\mathrm{T}^{2}$
<br/> <br/>$$
<br/> <br/>\begin{array}{l}
<br/> <br/>\text { or } \quad \mathrm{S}=\mathrm{x}^{2}+\mathrm{y}^{2}+20(\mathrm{x}+\mathrm{y})+20, \mathrm{~S}_{1}=20 \\
<br/> <br/>\mathrm{~T}=10(\mathrm{x}+\mathrm{y})+20=0 \\
<br/> <br/>\therefore \quad \mathrm{SS}_{1}=\mathrm{T}^{2} \\
<br/> <br/>\Rightarrow \quad 20\left(\mathrm{x}^{2}+\mathrm{y}^{2}+20(\mathrm{x}+\mathrm{y})+20\right)=10^{2} \\
<br/> <br/>\quad(\mathrm{x}+\mathrm{y}+2)^{2} \\
<br/> <br/>\Rightarrow \quad 4 \mathrm{x}^{2}+4 \mathrm{y}^{2}+10 \mathrm{xy}=0 \Rightarrow 2 \mathrm{x}^{2}+2 \mathrm{y}^{2}+5 \mathrm{xy}=0
<br/> <br/>\end{array}
<br/> <br/>$$</div>


tags: ['Physics', 'Capacitance', 'BITSAT', 'BITSAT 2018']
source_db: MarksBatch1_P1.db
title:
a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-d-3-i

slug:
a-parallel-plate-capacitor-c-with-plates-of-unit-area-and-separation-d-is-filled-with-a-liquid-of-dielectric-constant-k-2-the-level-of-liquid-is-d-3-i-12244

question:
<div class="question"><p>A parallel plate capacitor <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi></math> with plates of unit area and separation <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math> is filled with a liquid of dielectric constant <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi><mo>=</mo><mn>2</mn></math>, the level of liquid is<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>d</mi><mn>3</mn></mfrac><mo>,</mo></math>initially. <img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/974c2e8a-cf06-4938-97ce-f4fba41c1918-image.png" style="width: 300px; height: 183px;"/><br/>Suppose, the liquid level decreases at a constant speed <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi></math>, the time constant as a function of time is</p></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>6</mn><msub><mi>ε</mi><mn>0</mn></msub><mi>R</mi></mrow><mrow><mn>5</mn><mi>d</mi><mo>+</mo><mn>3</mn><mi>v</mi><mi>t</mi></mrow></mfrac></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mo>(</mo><mn>15</mn><mi>d</mi><mo>+</mo><mn>9</mn><mi>v</mi><mi>t</mi><mo>)</mo><msub><mi>ε</mi><mn>0</mn></msub><mi>R</mi></mrow><mrow><mn>2</mn><msup><mi>d</mi><mn>3</mn></msup><mo>-</mo><mn>3</mn><mi>d</mi><mi>v</mi><mi>t</mi><mo>-</mo><mi>g</mi><msup><mi>v</mi><mn>2</mn></msup><msup><mi>t</mi><mn>2</mn></msup></mrow></mfrac></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>6</mn><msub><mi>ε</mi><mn>0</mn></msub><mi>R</mi></mrow><mrow><mn>5</mn><mi>d</mi><mo>-</mo><mn>3</mn><mi>v</mi><mi>t</mi></mrow></mfrac></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mo>(</mo><mn>15</mn><mi>d</mi><mo>-</mo><mn>9</mn><mi>v</mi><mi>t</mi><mo>)</mo><msub><mi>ε</mi><mn>0</mn></msub><mi>R</mi></mrow><mrow><mn>2</mn><msup><mi>d</mi><mn>3</mn></msup><mo>+</mo><mn>3</mn><mi>d</mi><mi>v</mi><mi>t</mi><mo>-</mo><mn>9</mn><msup><mi>v</mi><mn>2</mn></msup><msup><mi>t</mi><mn>2</mn></msup></mrow></mfrac></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>6</mn><msub><mi>ε</mi><mn>0</mn></msub><mi>R</mi></mrow><mrow><mn>5</mn><mi>d</mi><mo>+</mo><mn>3</mn><mi>v</mi><mi>t</mi></mrow></mfrac></math></span> </div>

answer:
<div class="solution"><p>We have, the time constant, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>τ</mi><mo>=</mo><mi>R</mi><mi>C</mi><mo>'</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>…</mo><mfenced><mi mathvariant="normal">i</mi></mfenced></math></p><p>Now, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>'</mo><mo>=</mo><mfrac><mrow><msub><mi>C</mi><mn>1</mn></msub><msub><mi>C</mi><mn>2</mn></msub></mrow><mrow><msub><mi>C</mi><mn>1</mn></msub><mo>+</mo><msub><mi>C</mi><mn>2</mn></msub></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mrow><mfenced><mfrac><mrow><mi>A</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mrow><mi>d</mi><mo>-</mo><mi>x</mi></mrow></mfrac></mfenced><mfenced><mfrac><mrow><mi>K</mi><mi>A</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mi>x</mi></mfrac></mfenced></mrow><mrow><mfrac><mrow><mi>A</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mrow><mi>d</mi><mo>-</mo><mi>x</mi></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>K</mi><mi>A</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mi>x</mi></mfrac></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo>  </mo><mi>C</mi><mo>'</mo><mo>=</mo><mfrac><mrow><mi>K</mi><mi>A</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mrow><mi>x</mi><mo>+</mo><mi>K</mi><mo>(</mo><mi>d</mi><mo>-</mo><mi>x</mi><mo>)</mo></mrow></mfrac></math></p><p>Putting the value of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>'</mo></math> in Equation. (i), we get</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>τ</mi><mo>=</mo><mfrac><mrow><mi>R</mi><mi>K</mi><mi>A</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mrow><mfrac><mi>d</mi><mn>3</mn></mfrac><mo>-</mo><mi>v</mi><mi>t</mi><mo>+</mo><mi>k</mi><mfenced><mrow><mi>d</mi><mo>-</mo><mfrac><mi>d</mi><mn>3</mn></mfrac><mo>+</mo><mi>v</mi><mi>t</mi></mrow></mfenced></mrow></mfrac><mfenced close="]" open="["><mrow><mo>∵</mo><mi>x</mi><mo>=</mo><mfrac><mi>d</mi><mn>3</mn></mfrac><mo>-</mo><mi>v</mi><mi>t</mi></mrow></mfenced></math></p><p>Given, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi><mo>=</mo><mn>1</mn></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi><mo>=</mo><mn>2</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo>  </mo><mi>τ</mi><mo>=</mo><mfrac><mrow><mn>3</mn><mo>×</mo><mn>2</mn><msub><mi>ε</mi><mn>0</mn></msub><mo>×</mo><mi>R</mi></mrow><mrow><mi>d</mi><mo>-</mo><mn>3</mn><mi>v</mi><mi>t</mi><mo>+</mo><mn>6</mn><mi>d</mi><mo>-</mo><mn>2</mn><mi>d</mi><mo>+</mo><mn>6</mn><mi>v</mi><mi>t</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>6</mn><mi>R</mi><msub><mi>ε</mi><mn>0</mn></msub></mrow><mrow><mn>5</mn><mi>d</mi><mo>+</mo><mn>3</mn><mi>v</mi><mi>t</mi></mrow></mfrac></math></p></div>


tags: ['Physics', 'Capacitance', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-parallel-plate-capacitor-of-area-a-plate-separation-d-is-filled-with-two-dielectrics-as-shown-what-is-the-capacitance-of-the-arrangement-1

slug:
a-parallel-plate-capacitor-of-area-a-plate-separation-d-is-filled-with-two-dielectrics-as-shown-what-is-the-capacitance-of-the-arrangement-1-28865

question:
<div class="question">A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/Uwdipp_XL3YUGOKXcHsavEPIHKMk-pr-KF77AodoFoU.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{3 K \varepsilon_{0} A}{4 d}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{4 K \varepsilon_{0} A}{3 d}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{(K+1) \varepsilon_{0} A}{2 d}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$</span> </div>

answer:
<div class="solution">$$
<br/> <br/>c_{1}=\frac{(A / 2) \varepsilon_{0}}{d / 2}=\frac{A \varepsilon_{0}}{d}, c_{2}=K \frac{A \varepsilon_{0}}{d}, c_{3}=K \frac{A \varepsilon_{0}}{2 d}
<br/> <br/>$$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/-2J2c5Qbdiu4RPjNxqVYEm9IxIFFFQQ25PfTCZIJjOo.original.fullsize.png"/><br/> <br/> <br/>$$
<br/> <br/>\therefore \quad c_{e q .}=\frac{c_{1} \times c_{2}}{c_{1}+c_{2}}+c_{3}=\frac{(3+\mathrm{K}) \mathrm{K} A \varepsilon_{0}}{2 d(\mathrm{~K}+1)}
<br/> <br/>$$
<br/> <br/>$\left(\because C_{1}\right.$ and $C_{2}$ are in series and resultant of these two in parallel with $C_{3}$ )</div>


tags: ['Physics', 'Capacitance', 'BITSAT', 'BITSAT 2016']
source_db: MarksBatch1_P1.db
title:
a-parallel-plate-capacitor-of-area-a-plate-separation-d-is-filled-with-two-dielectrics-as-shown-what-is-the-capacitance-of-the-arrangement

slug:
a-parallel-plate-capacitor-of-area-a-plate-separation-d-is-filled-with-two-dielectrics-as-shown-what-is-the-capacitance-of-the-arrangement-48048

question:
<div class="question">A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/Uwdipp_XL3YUGOKXcHsavEPIHKMk-pr-KF77AodoFoU.original.fullsize.png"/></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{3 K \varepsilon_{0} A}{4 d}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{4 K \varepsilon_{0} A}{3 d}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{(K+1) \varepsilon_{0} A}{2 d}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{K(K+3) \varepsilon_{0} A}{2(K+1) d}$</span> </div>

answer:
<div class="solution">$$
<br/> <br/>c_{1}=\frac{(A / 2) \varepsilon_{0}}{d / 2}=\frac{A \varepsilon_{0}}{d}, c_{2}=K \frac{A \varepsilon_{0}}{d}, c_{3}=K \frac{A \varepsilon_{0}}{2 d}
<br/> <br/>$$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/-2J2c5Qbdiu4RPjNxqVYEm9IxIFFFQQ25PfTCZIJjOo.original.fullsize.png"/><br/> <br/> <br/>$$
<br/> <br/>\therefore \quad c_{e q .}=\frac{c_{1} \times c_{2}}{c_{1}+c_{2}}+c_{3}=\frac{(3+\mathrm{K}) \mathrm{K} A \varepsilon_{0}}{2 d(\mathrm{~K}+1)}
<br/> <br/>$$
<br/> <br/>$\left(\because C_{1}\right.$ and $C_{2}$ are in series and resultant of these two in parallel with $C_{3}$ )</div>


tags: ['Physics', 'Capacitance', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-parallel-plate-capacitor-of-capacitance-c-is-connected-to-a-battery-and-is-charged-to-a-potential-difference-v-another-capacitor-of-capacitance-2-c--1

slug:
a-parallel-plate-capacitor-of-capacitance-c-is-connected-to-a-battery-and-is-charged-to-a-potential-difference-v-another-capacitor-of-capacitance-2-c-1-52825

question:
<div class="question">A parallel plate capacitor of capacitance $\mathrm{C}$ is connected to a battery and is charged to a potential difference $\mathrm{V}$. Another capacitor of capacitance $2 \mathrm{C}$ is similarly charge to a potential difference $2 \mathrm{~V}$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Zero</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\frac{3}{2} \mathrm{CV}^{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{25}{6} \mathrm{CV}^{2}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{9}{2} \mathrm{CV}^{2}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{3}{2} \mathrm{CV}^{2}$</span> </div>

answer:
<div class="solution">From the figure. The net charge shared between the capacitors
<br/> <br/>$$
<br/> <br/>Q^{\prime}=Q_{2}-Q_{2}=4 C V-C V=3 C V
<br/> <br/>$$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/uC14fhROzIYYJ0pa2wGnML5VYknjH1G83FGfD57wcRo.original.fullsize.png"/><br/> <br/> <br/>The two capacitors will have some potential, say $\mathrm{V}^{\prime}$. The net capacitance of the parallel combination of the two capacitors
<br/> <br/>$$
<br/> <br/>C^{\prime}=C_{1}+C_{2}=C+2 C+3 C
<br/> <br/>$$
<br/> <br/>The potential of the capacitors
<br/> <br/>$$
<br/> <br/>\mathrm{V}^{\prime}=\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{3 \mathrm{CV}}{3 \mathrm{C}}=\mathrm{V}
<br/> <br/>$$
<br/> <br/>The electrostatic energy of the capacitors
<br/> <br/>$$
<br/> <br/>\mathrm{E}^{\prime}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{\prime 2}=\frac{1}{2}(3 \mathrm{C}) \mathrm{V}^{2}=\frac{3}{2} \mathrm{CV}^{2}
<br/> <br/>$$</div>


tags: ['Physics', 'Capacitance', 'BITSAT', 'BITSAT 2017']
source_db: MarksBatch1_P1.db
title:
a-parallel-plate-capacitor-with-air-between-the-plates-has-a-capacitance-of-9-pf-the-separation-between-the-plates-is-d-the-space-between-the-plates-i

slug:
a-parallel-plate-capacitor-with-air-between-the-plates-has-a-capacitance-of-9-pf-the-separation-between-the-plates-is-d-the-space-between-the-plates-i-45398

question:
<div class="question">A parallel plate capacitor with air between the plates has a capacitance of <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>9</mn><mo> </mo><mi>pF</mi></math>. The separation between the plates is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi></math>. The space between the plates is now filled with two dielectrics constant <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>K</mi><mn>1</mn></msub><mo>=</mo><mn>3</mn></math> and thickness <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>d</mi><mn>3</mn></mfrac></math> while the other one has dielectric constant <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>K</mi><mn>2</mn></msub><mo>=</mo><mn>6</mn></math> and thickness <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>2</mn><mi>d</mi></mrow><mn>3</mn></mfrac></math>. Capacitance of the capacitor is now </div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>.</mo><mn>8</mn><mo> </mo><mi>pF</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>45</mn><mo> </mo><mi>pF</mi></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>40</mn><mo>.</mo><mn>5</mn><mo> </mo><mi>pF</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn><mo>.</mo><mn>25</mn><mo> </mo><mi>pF</mi></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>40</mn><mo>.</mo><mn>5</mn><mo> </mo><mi>pF</mi></math></span> </div>

answer:
<div class="solution"><p>The two capacitors formed are in parallel, hence capacitance of the combination</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>=</mo><mfrac><mrow><msub><mi>C</mi><mn>1</mn></msub><msub><mi>C</mi><mn>2</mn></msub></mrow><mrow><msub><mi>C</mi><mn>1</mn></msub><mo>+</mo><msub><mi>C</mi><mn>2</mn></msub></mrow></mfrac></math>   <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>.</mo><mo>.</mo><mo>.</mo><mo>(</mo><mi mathvariant="normal">i</mi><mo>)</mo></math></p><p>where, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>1</mn></msub><mo>=</mo><mfrac><mrow><msub><mi>K</mi><mn>1</mn></msub><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi></mrow><mfrac><mi>d</mi><mn>3</mn></mfrac></mfrac></math>    <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mo>∵</mo><mi>C</mi><mo>=</mo><mfrac><mrow><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi></mrow><mi>d</mi></mfrac></mrow></mfenced></math>   <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>.</mo><mo>.</mo><mo>.</mo><mo>(</mo><mi>ii</mi><mo>)</mo></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>2</mn></msub><mo>=</mo><mfrac><mrow><msub><mi>K</mi><mn>2</mn></msub><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi></mrow><mfrac><mrow><mn>2</mn><mi>d</mi></mrow><mn>3</mn></mfrac></mfrac></math>    <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>.</mo><mo>.</mo><mo>.</mo><mo>(</mo><mi>iii</mi><mo>)</mo></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mi>eq</mi></msub><mo>=</mo><mfrac><mrow><mfrac><mrow><mn>3</mn><msub><mi>K</mi><mn>1</mn></msub><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi></mrow><mi>d</mi></mfrac><mo>×</mo><mfrac><mrow><msub><mi>K</mi><mn>2</mn></msub><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi><mo>·</mo><mn>3</mn></mrow><mrow><mn>2</mn><mi>d</mi></mrow></mfrac></mrow><mrow><mfrac><mrow><mn>3</mn><msub><mi>K</mi><mn>1</mn></msub><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi></mrow><mi>d</mi></mfrac><mo>+</mo><mfrac><mrow><mn>3</mn><msub><mi>K</mi><mn>2</mn></msub><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi></mrow><mrow><mn>2</mn><mi>d</mi></mrow></mfrac></mrow></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mrow><mn>9</mn><msub><mi>K</mi><mn>1</mn></msub><msub><mi>K</mi><mn>2</mn></msub><msubsup><mi>ε</mi><mn>0</mn><mn>2</mn></msubsup><msup><mi>A</mi><mn>2</mn></msup></mrow><mrow><mn>2</mn><msup><mi>d</mi><mn>2</mn></msup></mrow></mfrac><mo>×</mo><mfrac><mi>d</mi><mrow><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi><mo>×</mo><mn>18</mn></mrow></mfrac></math></p><p>It is given</p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/5b6eb6e7-1086-4057-8b85-750395666a01-image.png" style="width: 210px; height: 123px;"/></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><msub><mi>ε</mi><mn>0</mn></msub><mi>A</mi></mrow><mi>d</mi></mfrac><mo>=</mo><mn>9</mn><mo> </mo><mi>pF</mi></math></p><p>Using given values</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mi>eq</mi></msub><mo>=</mo><mn>40</mn><mo>.</mo><mn>5</mn><mo> </mo><mi>pF</mi></math></p></div>


tags: ['Physics', 'Oscillations', 'BITSAT', 'BITSAT 2023 (Memory Based Paper 1)']
source_db: MarksBatch1_P1.db
title:
a-particle-executes-shm-of-amplitude-a-the-distance-from-the-mean-position-when-its-kinetic-energy-becomes-equal-to-its-potential-energy-is-1

slug:
a-particle-executes-shm-of-amplitude-a-the-distance-from-the-mean-position-when-its-kinetic-energy-becomes-equal-to-its-potential-energy-is-1-92116

question:
<div class="question">A particle executes SHM of amplitude A. The distance from the mean position when it's kinetic energy becomes equal to its potential energy is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\sqrt{2 \mathrm{~A}}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$2 \mathrm{~A}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\frac{1}{\sqrt{2}} \mathrm{~A}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{1}{2} \mathrm{~A}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{1}{\sqrt{2}} \mathrm{~A}$</span> </div>

answer:
<div class="solution">Let the distance from the mean position is $\mathrm{X}$.<br/>Given $\mathrm{KE}=\mathrm{PE}$<br/>So, $\frac{1}{2} M \omega^2\left(A^2-x^2\right)=\frac{1}{2} M \omega^2 x^2$<br/>$A^2-x^2=x^2 \Rightarrow A^2=2 \times 2$<br/>$\therefore \mathrm{x}= \pm \frac{\mathrm{A}}{\sqrt{2}}$</div>


tags: ['Physics', 'Oscillations', 'BITSAT', 'BITSAT 2020']
answer: <div class="solution">The amplitude is a maximum displacement from the mean position.</div>
source_db: MarksBatch1_P1.db
title:
a-particle-executing-simple-harmonic-motion-along-y-axis-has-its-motion-described-by-the-equation-y-a-sin-t-b-the-amplitude-of-the-simple-harmonic-mot

slug:
a-particle-executing-simple-harmonic-motion-along-y-axis-has-its-motion-described-by-the-equation-y-a-sin-t-b-the-amplitude-of-the-simple-harmonic-mot-57728

question:
<div class="question">A particle executing simple harmonic motion along $y$ -axis has its motion described by the equation $y=A \sin (\omega t)+B$. The amplitude of the simple harmonic motion is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\mathrm{A}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{B}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{A}+\mathrm{B}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\sqrt{A+B}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{A}$</span> </div>


tags: ['Physics', 'Motion In One Dimension', 'BITSAT', 'BITSAT 2011']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">50</span> </div>
answer: <div class="solution">No Solution available</div>
source_db: MarksBatch1_P1.db
title:
a-particle-is-moving-in-a-straight-line-with-initial-velocity-and-uniform-acceleration-a-if-the-sum-of-the-distance-travelled-in-t-th-and-t-1-th-secon

slug:
a-particle-is-moving-in-a-straight-line-with-initial-velocity-and-uniform-acceleration-a-if-the-sum-of-the-distance-travelled-in-t-th-and-t-1-th-secon-21677

question:
<div class="question">A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in $t^{\text {th }}$ and $(t+1)^{\text {th }}$ seconds is $100 \mathrm{~cm}$, then its velocity after $t$ seconds, in $\mathrm{cm} / \mathrm{s}, \mathrm{is}$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">80</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">50</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">20</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">30</span> </li> </ul>


tags: ['Physics', 'Motion In One Dimension', 'BITSAT', 'BITSAT 2017']
source_db: MarksBatch1_P1.db
title:
a-particle-moving-along-x-axis-has-acceleration-f-at-time-t-given-by-f-f-0-1-t-t-where-f-0-and-t-are-constants-the-particle-at-t-0-and-the-instant-whe

slug:
a-particle-moving-along-x-axis-has-acceleration-f-at-time-t-given-by-f-f-0-1-t-t-where-f-0-and-t-are-constants-the-particle-at-t-0-and-the-instant-whe-64218

question:
<div class="question">A particle moving along <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>X</mi></math>-axis has acceleration <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi></math>, at time <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi></math> given by<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><msub><mi>f</mi><mn>0</mn></msub><mfenced><mrow><mn>1</mn><mo>-</mo><mfrac><mi>t</mi><mi>T</mi></mfrac></mrow></mfenced></math>, where <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>f</mi><mn>0</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi></math> are constants. The particle at <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mn>0</mn></math> and the instant when <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mn>0</mn></math>, the particle's velocity <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>v</mi><mi>x</mi></msub></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>f</mi><mn>0</mn></msub><msup><mi>T</mi><mn>2</mn></msup></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>f</mi><mn>0</mn></msub><msup><mi>T</mi><mn>2</mn></msup></math></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi></math></span> </div>

answer:
<div class="solution"><p> Acceleration</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mfrac><mrow><mi>d</mi><mi>v</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>=</mo><msub><mi>f</mi><mn>0</mn></msub><mfenced><mrow><mn>1</mn><mo>+</mo><mfrac><mi>t</mi><mi>T</mi></mfrac></mrow></mfenced></math></p><p>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mi>v</mi><mo>=</mo><msub><mi>f</mi><mn>0</mn></msub><mfenced><mrow><mn>1</mn><mo>-</mo><mfrac><mi>t</mi><mi>T</mi></mfrac></mrow></mfenced><mo>·</mo><mi>d</mi><mi>t</mi><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mo> </mo><mi>E</mi><mi>q</mi><mo>(</mo><mi>i</mi><mo>)</mo></math></p><p>Integrating Eq. <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mi mathvariant="normal">i</mi><mo>)</mo></math> on both sides, we get</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>=</mo><msub><mi>f</mi><mn>0</mn></msub><mi>t</mi><mo>-</mo><mfrac><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi></mfrac><mo>·</mo><mfrac><msup><mi>t</mi><mn>2</mn></msup><mn>2</mn></mfrac><mo>+</mo><mi>C</mi></math></p><p>After applying boundary conditions </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>v</mi><mo>=</mo><mn>0</mn></math> at <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mn>0</mn></math></p><p>We get, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mi>v</mi><mo>=</mo><msub><mi>f</mi><mn>0</mn></msub><mi>t</mi><mo>-</mo><mfrac><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi></mfrac><mo>·</mo><mfrac><msup><mi>t</mi><mn>2</mn></msup><mn>2</mn></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo> </mo><mo> </mo><mi>E</mi><mi>q</mi><mo>(</mo><mi>i</mi><mi>i</mi><mo>)</mo></math>   </p><p>As <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><msub><mi>f</mi><mn>0</mn></msub><mfenced><mrow><mn>1</mn><mo>-</mo><mfrac><mi>t</mi><mi>T</mi></mfrac></mrow></mfenced></math></p><p>When <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>f</mi><mn>0</mn></msub><mo>=</mo><mn>0</mn><mo>,</mo><mi>t</mi><mo>=</mo><mi>T</mi></math></p><p>Substituting <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mi>T</mi></math> in Eq. <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mi>ii</mi><mo>)</mo></math>, then velocity</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>v</mi><mi>x</mi></msub><mo>=</mo><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi><mo>-</mo><mfrac><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi></mfrac><mo>·</mo><mfrac><msup><mi>T</mi><mn>2</mn></msup><mn>2</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><msub><mi>f</mi><mn>0</mn></msub><mi>T</mi></math></p></div>


tags: ['Physics', 'Motion In Two Dimensions', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-2-m-is-projected-at-an-angle-of-4-5-with-the-horizontal-with-a-velocity-of-20-2-m-s-after-1-s-explosion-takes-place-and-the-particl

slug:
a-particle-of-mass-2-m-is-projected-at-an-angle-of-4-5-with-the-horizontal-with-a-velocity-of-20-2-m-s-after-1-s-explosion-takes-place-and-the-particl-90081

question:
<div class="question">A particle of mass $2 \mathrm{~m}$ is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $20 \sqrt{2} \mathrm{~m} / \mathrm{s}$. After $1 \mathrm{~s}$, explosion takes place and the particle is broken into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$50 \mathrm{~m}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$25 \mathrm{~m}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$40 \mathrm{~m}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$35 \mathrm{~m}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$35 \mathrm{~m}$</span> </div>

answer:
<div class="solution">Given : Initial velocity $u_{0}=20 \sqrt{2} \mathrm{~m} / \mathrm{s} ;$ angle of projection $\theta=45^{\circ}$
<br/> <br/>Therefore horizontal and vertical components of initial velocity are $\mathrm{u}_{\mathrm{x}}=20 \sqrt{2} \cos 45^{\circ}=20 \mathrm{~m} / \mathrm{s}$
<br/> <br/>and $\mathrm{u}_{\mathrm{y}}=20 \sqrt{2} \sin 45^{\circ}=20 \mathrm{~m} / \mathrm{s}$
<br/> <br/>After $1 \mathrm{~s}$, horizontal component remains unchanged while the vertical component becomes $v_{y}=u_{y}-g t$
<br/> <br/>Due to explosion, one part comes to rest. Hence, from the conservation of linear momentum, vertical component of second
<br/> <br/>part will become $v_{y}^{\prime}=20 m / s$ Therefore, maximum height attained by the second part will be $\mathrm{H}=\mathrm{h}_{1}+\mathrm{h}_{2}$
<br/> <br/>Using, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$
<br/> <br/>$\Rightarrow \mathrm{h}_{1}=(20 \times 1)-\frac{1}{2} \times 10 \times(1)^{2}=15 \mathrm{~m}$
<br/> <br/>$\mathrm{a}=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$
<br/> <br/>$\mathrm{h}_{2}=\frac{\mathrm{v}_{\mathrm{y}}^{\prime 2}}{2 \mathrm{~g}}=\frac{(20)^{2}}{2 \times 10}=20 \mathrm{~m}$
<br/> <br/>$\mathrm{H}=20+15=35 \mathrm{~m}$</div>


tags: ['Physics', 'Center of Mass Momentum and Collision', 'BITSAT', 'BITSAT 2009']
source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-10-kg-moving-eastwards-with-a-speed-5-ms-1-collides-with-another-particle-of-the-same-mass-moving-northwards-with-the-same-speed-5-

slug:
a-particle-of-mass-10-kg-moving-eastwards-with-a-speed-5-ms-1-collides-with-another-particle-of-the-same-mass-moving-northwards-with-the-same-speed-5-79732

question:
<div class="question">A particle of mass $10 \mathrm{~kg}$ moving eastwards with a speed $5 \mathrm{~ms}^{-1}$ collides with another particle of the same mass moving north-wards with the same speed $5 \mathrm{~ms}^{-1}$. The two particles coalesce on collision. The new particle of mass $20 \mathrm{~kg}$ will move in the north-east direction with velocity</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$10 \mathrm{~ms}^{-1}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$5 \mathrm{~ms}^{-1}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\quad(5 / \sqrt{2}) \mathrm{ms}^{-1}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">none of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\quad(5 / \sqrt{2}) \mathrm{ms}^{-1}$</span> </div>

answer:
<div class="solution">Here $\hat{i} m v+\hat{j} m v=2 m \vec{V}$<br/>That is $\overrightarrow{\mathrm{V}}=\frac{\mathrm{v}}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$<br/>Hence $V=\frac{v}{2} \times \sqrt{2}=\frac{v}{\sqrt{2}} \cdot\left[\right.$ Here $\left.v=5 \mathrm{~ms}^{-1}\right]$<br/>$$<br/>\text { So, } V=\frac{5}{\sqrt{2}} \mathrm{~ms}^{-1}<br/>$$</div>


tags: ['Physics', 'Rotational Motion', 'JEE Main']
source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-m-is-projected-with-a-velocity-v-making-an-angle-of-3-0-with-the-horizontal-the-magnitude-of-angular-momentum-of-the-projectile-abo-1

slug:
a-particle-of-mass-m-is-projected-with-a-velocity-v-making-an-angle-of-3-0-with-the-horizontal-the-magnitude-of-angular-momentum-of-the-projectile-abo-1-15659

question:
<div class="question">A particle of mass ' $m$ ' is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ' $h$ ' is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{\sqrt{3}}{2} \frac{m v^{2}}{g}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">zero</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{m v^{3}}{\sqrt{2 g}}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$</span> </div>

answer:
<div class="solution">Angular momentum $L_{0}=p r_{\perp}$ $\left(\because\right.$ linear momentum $p=m v \cos \theta$ and $\left.\mathrm{r}_{1}=H\right)$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/nZ84PuGwoTRtrwb2Py798AbqUV7RMNwWKZjAOgiqSiM.original.fullsize.png"/><br/> <br/> <br/>$\Rightarrow L_{0}=m v \cos \theta H$
<br/> <br/>$=m v \frac{\sqrt{3}}{2} \cdot \frac{v^{2} \sin ^{2} 30^{\circ}}{2 g}$
<br/> <br/>$=\frac{\sqrt{3} m v^{3}}{16 g}$</div>


tags: ['Physics', 'Rotational Motion', 'BITSAT', 'BITSAT 2013']
source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-m-is-projected-with-a-velocity-v-making-an-angle-of-3-0-with-the-horizontal-the-magnitude-of-angular-momentum-of-the-projectile-abo

slug:
a-particle-of-mass-m-is-projected-with-a-velocity-v-making-an-angle-of-3-0-with-the-horizontal-the-magnitude-of-angular-momentum-of-the-projectile-abo-62439

question:
<div class="question">A particle of mass ' $m$ ' is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ' $h$ ' is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{\sqrt{3}}{2} \frac{m v^{2}}{g}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">zero</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{m v^{3}}{\sqrt{2 g}}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$</span> </div>

answer:
<div class="solution">Angular momentum $L_{0}=p r_{\perp}$ $\left(\because\right.$ linear momentum $p=m v \cos \theta$ and $\left.\mathrm{r}_{1}=H\right)$
<br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/nZ84PuGwoTRtrwb2Py798AbqUV7RMNwWKZjAOgiqSiM.original.fullsize.png"/><br/> <br/> <br/>$\Rightarrow L_{0}=m v \cos \theta H$
<br/> <br/>$=m v \frac{\sqrt{3}}{2} \cdot \frac{v^{2} \sin ^{2} 30^{\circ}}{2 g}$
<br/> <br/>$=\frac{\sqrt{3} m v^{3}}{16 g}$</div>


tags: ['Physics', 'Laws of Motion', 'BITSAT', 'BITSAT 2022']
source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-m-is-projected-with-velocity-v-at-an-angle-with-the-horizontal-at-its-highest-point-it-explodes-into-two-pieces-of-equal-mass-one-o

slug:
a-particle-of-mass-m-is-projected-with-velocity-v-at-an-angle-with-the-horizontal-at-its-highest-point-it-explodes-into-two-pieces-of-equal-mass-one-o-31730

question:
<div class="question">A particle of mass $m$ is projected with velocity $v$ at an angle $\theta$ with the horizontal. At its highest point, it explodes into two pieces of equal mass, one of the piece continue to move on the original trajectory, then the velocity of second piece is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"> <br/>$2 v \cos \theta$
<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"> <br/>$v \cos \theta$
<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"> <br/>$3 v \cos \theta$
<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"> <br/>$\frac{v}{2} \cos \theta$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"> <br/>$v \cos \theta$
<br/></span> </div>

answer:
<div class="solution">According to the law of conservation of linear momentum at the highest point,
<br/>$$
<br/>m v \cos \theta=\frac{m}{2}(v \cos \theta)+\frac{m}{2} v^{\prime}
<br/>$$
<br/>$\Rightarrow \quad m v \cos \theta-\frac{m}{2} v \cos \theta=\frac{m}{2} v^{\prime}$
<br/>$\Rightarrow \quad v^{\prime}=2\left(v \cos \theta-\frac{v \cos \theta}{2}\right)=v \cos \theta$</div>


source_db: MarksBatch1_P1.db
title:
a-particle-of-mass-m-moving-with-velocity-v-collides-with-a-stationary-particle-of-mass-2-m-after-collision-they-stick-together-and-continue-to-move-t-1

slug:
a-particle-of-mass-m-moving-with-velocity-v-collides-with-a-stationary-particle-of-mass-2-m-after-collision-they-stick-together-and-continue-to-move-t-1-18779

question:
<div class="question">A particle of mass $m$ moving with velocity $v$ collides with a stationary particle of mass $2 \mathrm{~m}$. After collision, they stick together and continue to move together with velocity</div>

tags:
['Physics', 'Center of Mass Momentum and Collision', 'BITSAT', 'BITSAT 2023 (Memory Based Paper 1)']

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{v}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{\mathrm{v}}{2}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\frac{\mathrm{v}}{3}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{v}{4}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{\mathrm{v}}{3}$</span> </div>

answer:
<div class="solution"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/SbGKuYro4oi5MfE5YaI3FNs5wpXRnpmNdsMqjZHtRo0.original.fullsize.png"/><br/><br/>Applying conservation of linear momentum<br/>$\Rightarrow \overrightarrow{\mathrm{P}}_{\mathrm{i}}=\overrightarrow{\mathrm{P}}_{\mathrm{f}}(\because \mathrm{P}=\mathrm{mv})$<br/>$m v_1+2 m v_2=(m+2 m) v^{\prime}$<br/>$m v^1+2 m \times 0=(3 m) v^{\prime}$<br/>$\Rightarrow \mathrm{mv}=3 \mathrm{mv}^{\prime} \Rightarrow \mathrm{v}^{\prime}=\frac{\mathrm{v}}{3}$</div>


tags: ['Physics', 'Motion In One Dimension', 'BITSAT', 'BITSAT 2009']
source_db: MarksBatch1_P1.db
title:
a-particle-starts-moving-rectilinearly-at-time-t-0-such-that-its-velocity-v-changes-with-time-t-according-to-the-equation-v-t-2-t-where-t-is-in-second

slug:
a-particle-starts-moving-rectilinearly-at-time-t-0-such-that-its-velocity-v-changes-with-time-t-according-to-the-equation-v-t-2-t-where-t-is-in-second-37720

question:
<div class="question">A particle starts moving rectilinearly at time $\mathrm{t}=0$ such that its velocity $\mathrm{v}$ changes with time $\mathrm{t}$ according to the equation $v=t^{2}-t$ where $t$ is in seconds and $v$ is in $\mathrm{m} / \mathrm{s}$. Find the time interval for which the particle retards.</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\frac{1}{2}&lt;\mathrm{t}&lt;1$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{1}{2}&gt;t&gt;1$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{1}{4}&lt;\mathrm{t}&lt;1$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{1}{2}<t> </t></span></li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{1}{2}&lt;\mathrm{t}&lt;1$</span> </div>

answer:
<div class="solution">Acceleration of the particle $a=2 t-1$ The particle retards when acceleration is opposite to velocity.<br/>$$<br/>\Rightarrow \text { a } \cdot v&lt;0 \Rightarrow(2 t-1)\left(t^{2}-t\right)&lt;0 \Rightarrow t(2 t-1)<br/>$$<br/>$(\mathrm{t}-1)&lt;0$<br/>Now t is always positive<br/>$$<br/>\therefore(2 t-1)(t-1)&lt;0<br/>$$<br/>or $2 \mathrm{t}-1&lt;0$ and $\mathrm{t}-1&gt;0 \Rightarrow \mathrm{t}&lt;\frac{1}{2}$ and $\mathrm{t}&gt;1$.<br/>This is not possible or $2 \mathrm{t}-1&gt;0 \&amp; \mathrm{t}-1&lt;0 \Rightarrow 1 / 2&lt;\mathrm{t}&lt;1$</div>


tags: ['Physics', 'Dual Nature of Matter', 'BITSAT', 'BITSAT 2019']
source_db: MarksBatch1_P1.db
title:
a-particle-x-of-mass-m-and-initial-velocity-u-collides-with-another-particle-y-of-mass-3-m-4-which-is-at-rest-the-collision-is-headon-and-perfectly-el

slug:
a-particle-x-of-mass-m-and-initial-velocity-u-collides-with-another-particle-y-of-mass-3-m-4-which-is-at-rest-the-collision-is-headon-and-perfectly-el-63419

question:
<div class="question">A particle <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>X</mi></math> of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi></math> and initial velocity <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>u</mi></math> collides with another particle <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Y</mi></math> of mass <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>3</mn><mi>m</mi></mrow><mn>4</mn></mfrac></math>, which is at rest. The collision is head-on and perfectly elastic. The ratio of de-Broglie wavelengths <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>λ</mi><mi>Y</mi></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>λ</mi><mi>X</mi></msub></math> after the collision is,</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mo>:</mo><mn>3</mn></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo>:</mo><mn>32</mn></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>:</mo><mn>4</mn></math></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>:</mo><mn>6</mn></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mo>:</mo><mn>6</mn></math></span> </div>

answer:
<div class="solution"><p>For perfectly elastic collision,</p><p>total momentum before collision <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo></math> total momentum after collision</p><p>i.e. <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>u</mi><mo>+</mo><mfrac><mrow><mn>3</mn><mi>m</mi></mrow><mn>4</mn></mfrac><mo>×</mo><mn>0</mn><mo>=</mo><mi>m</mi><msub><mi>v</mi><mi>x</mi></msub><mo>+</mo><mfrac><mrow><mn>3</mn><mi>m</mi></mrow><mn>4</mn></mfrac><msub><mi>v</mi><mi>y</mi></msub></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mn>4</mn><mi>u</mi><mo>=</mo><mn>4</mn><msub><mi>v</mi><mi>x</mi></msub><mo>+</mo><mn>3</mn><msub><mi>v</mi><mi>y</mi></msub><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mi mathvariant="normal">i</mi></mfenced></math></p><p>Coefficient of restitution,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>e</mi><mo>=</mo><mfrac><mrow><mtext>Velocity of separation after collision</mtext><mo> </mo></mrow><mrow><mtext>Velocity of approach before collision</mtext><mo> </mo></mrow></mfrac><mo>=</mo><mfrac><mrow><mi> </mi><mfenced><mrow><msub><mi>v</mi><mi>y</mi></msub><mo>-</mo><msub><mi>v</mi><mi>x</mi></msub></mrow></mfenced></mrow><mfenced><mrow><msub><mi>u</mi><mi>x</mi></msub><mo>-</mo><msub><mi>u</mi><mi>y</mi></msub></mrow></mfenced></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mn>1</mn><mo>=</mo><mfrac><mrow><msub><mi>v</mi><mi>y</mi></msub><mo>-</mo><msub><mi>v</mi><mi>x</mi></msub></mrow><mrow><mi>u</mi><mo>-</mo><mn>0</mn></mrow></mfrac></math>   [<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∵</mo><mo> </mo><msub><mi>u</mi><mi>x</mi></msub><mo>=</mo><mi>u</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>u</mi><mi>y</mi></msub><mo>=</mo><mn>0</mn></math>]</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>u</mi><mo>=</mo><msub><mi>v</mi><mi>y</mi></msub><mo>-</mo><msub><mi>v</mi><mi>x</mi></msub><mo> </mo><mo> </mo><mo> </mo><mo>.</mo><mo>.</mo><mo>.</mo><mfenced><mi>ii</mi></mfenced></math></p><p>From equation (i) and (ii), we get</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>v</mi><mi>x</mi></msub><mo>=</mo><mfrac><mi>u</mi><mn>7</mn></mfrac></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>v</mi><mi>y</mi></msub><mo>=</mo><mfrac><mrow><mn>8</mn><mi>u</mi></mrow><mn>7</mn></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mfrac><msub><mi>λ</mi><mi>y</mi></msub><msub><mi>λ</mi><mi>x</mi></msub></mfrac><mo>=</mo><mfrac><mstyle displaystyle="true"><mfrac><mi>h</mi><mrow><mstyle displaystyle="true"><mfrac><mn>3</mn><mn>4</mn></mfrac></mstyle><mi>m</mi><msub><mi>v</mi><mi>y</mi></msub></mrow></mfrac></mstyle><mstyle displaystyle="true"><mfrac><mi>h</mi><mrow><mi>m</mi><mo>×</mo><msub><mi>v</mi><mi>x</mi></msub></mrow></mfrac></mstyle></mfrac><mo>=</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mo>×</mo><mfrac><msub><mi>v</mi><mi>x</mi></msub><msub><mi>v</mi><mi>y</mi></msub></mfrac><mo>=</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mfrac><mstyle displaystyle="true"><mfrac><mi>u</mi><mn>7</mn></mfrac></mstyle><mstyle displaystyle="true"><mfrac><mrow><mn>8</mn><mi>u</mi></mrow><mn>7</mn></mfrac></mstyle></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>6</mn></mfrac></math></p></div>


tags: ['Physics', 'Motion In Two Dimensions', 'BITSAT', 'BITSAT 2011']
source_db: MarksBatch1_P1.db
title:
a-passenger-in-a-open-car-travelling-at-30-m-s-throws-a-ball-out-over-the-bonnet-relative-to-the-car-the-initial-velocity-of-the-ball-is-20-m-s-at-6-0

slug:
a-passenger-in-a-open-car-travelling-at-30-m-s-throws-a-ball-out-over-the-bonnet-relative-to-the-car-the-initial-velocity-of-the-ball-is-20-m-s-at-6-0-57485

question:
<div class="question">A passenger in a open car travelling at $30 \mathrm{~m} / \mathrm{s}$ throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is $20 \mathrm{~m} / \mathrm{s}$ at $60^{\circ}$ to the horizontal. The angle of projection of the ball with respect to the horizontal road will be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\tan ^{-1}\left(\frac{2}{3}\right)$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\tan ^{-1}\left(\frac{\sqrt{3}}{4}\right)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\tan ^{-1}\left(\frac{4}{\sqrt{3}}\right)$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\tan ^{-1}\left(\frac{3}{4}\right)$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\tan ^{-1}\left(\frac{\sqrt{3}}{4}\right)$</span> </div>

answer:
<div class="solution">$$\begin{array}{l}<br/>\left(\overrightarrow{\mathrm{v}}_{\mathrm{bc}}\right)_{\mathrm{x}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}-\left(\overrightarrow{\mathrm{v}}_{\mathrm{c}}\right)_{\mathrm{x}} \\<br/>20 \cos 60^{\circ}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}-30 \\<br/>\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}=40 ;\left(\overrightarrow{\mathrm{v}}_{\mathrm{bc}}\right)_{\mathrm{y}}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}-\left(\overrightarrow{\mathrm{v}}_{\mathrm{c}}\right)_{\mathrm{y}} \\<br/>20 \sin 60^{\circ}=\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}-0 \\<br/>\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}=10 \sqrt{3} ; \tan \theta=\frac{\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{y}}}{\left(\overrightarrow{\mathrm{v}}_{\mathrm{b}}\right)_{\mathrm{x}}}=\frac{10 \sqrt{3}}{40}=\frac{\sqrt{3}}{4}<br/>\end{array}<br/>$$</div>


tags: ['Physics', 'Motion In One Dimension', 'BITSAT', 'BITSAT 2023 (Memory Based Paper 2)']
source_db: MarksBatch1_P1.db
title:
a-passenger-sitting-in-a-train-a-moving-at-90-km-h-observes-another-train-b-moving-in-the-opposite-direction-for-8-s-if-the-velocity-of-the-train-b-is

slug:
a-passenger-sitting-in-a-train-a-moving-at-90-km-h-observes-another-train-b-moving-in-the-opposite-direction-for-8-s-if-the-velocity-of-the-train-b-is-30422

question:
<div class="question">A passenger sitting in a train A moving at $90 \mathrm{~km} /$ $h$ observes another train B moving in the opposite direction for $8 \mathrm{~s}$. If the velocity of the train B is $54 \mathrm{~km} / \mathrm{h}$, then length of train B is :</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$80 \mathrm{~m}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$200 \mathrm{~m}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$120 \mathrm{~m}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$320 \mathrm{~m}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$320 \mathrm{~m}$</span> </div>

answer:
<div class="solution">Velocity of train A<br/>\(\mathrm{V}_{\mathrm{A}}=90 \frac{\mathrm{km}}{\mathrm{hr}}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)<br/>Velocity of train B<br/>\(\mathrm{V}_{\mathrm{B}}=54 \frac{\mathrm{km}}{\mathrm{hr}}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s}\)<br/>Velocity of train B w.r.t. train<br/>\(\begin{aligned}<br/>A=\vec{V}_B &amp; -\vec{V}_A \\<br/>&amp; =15-(-25) \mathrm{m} / \mathrm{s} \\<br/>&amp; =40 \mathrm{~m} / \mathrm{s}<br/>\end{aligned}\)<br/>Time of crossing \(=\frac{\text { length of train }}{\text { relative velocity }}\)<br/>\((8)=\frac{\ell}{40}\)<br/>\(\ell=8 \times 40=320\) meter.</div>


tags: ['Physics', 'Oscillations', 'BITSAT', 'BITSAT 2009']
source_db: MarksBatch1_P1.db
title:
a-pendulum-bob-carries-a-ve-charge-q-a-positive-charge-q-is-held-at-the-point-of-support-then-the-time-period-of-the-bob-is-where-l-length-of-pendulum

slug:
a-pendulum-bob-carries-a-ve-charge-q-a-positive-charge-q-is-held-at-the-point-of-support-then-the-time-period-of-the-bob-is-where-l-length-of-pendulum-92207

question:
<div class="question">A pendulum bob carries a +ve charge $+\mathrm{q} . \mathrm{A}$ positive charge $+\mathrm{q}$ is held at the point of support. Then the time period of the bob is [where, $\mathrm{L}=$ length of pendulum, $\mathrm{g}_{\mathrm{eff}}=$ effective value of $\mathrm{g}]$</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">greater than $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">less than $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">equal to $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">equal to $2 \pi \sqrt{2 \mathrm{~L} / \mathrm{g}_{\text {eff }}}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">greater than $2 \pi \sqrt{\mathrm{L} / \mathrm{g}_{\text {eff }}}$</span> </div>

answer:
<div class="solution">Effective $g^{\prime}=g-\frac{k q^{2}}{d^{2} m}, \quad T \propto \frac{1}{\sqrt{g_{\text {eff }}}}$</div>


title: a-penicillin-is-a-member-of-a-family-of-drugs-that-have-a
slug: a-penicillin-is-a-member-of-a-family-of-drugs-that-have-a-26795
question: <div class="question">A penicillin is a member of a family of drugs that have a</div>
tags: ['Chemistry', 'Chemistry in Everyday Life', 'BITSAT', 'BITSAT 2018']
source_db: MarksBatch1_P1.db
options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">four membered cyclic amide fused to a five membered thiazole ring.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">three membered cyclic amide fused to a five-membered thiazole ring.</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">four-membered cyclic amide fused to have a four membered thiazole ring.</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">five-membered cyclic amide fused to have a five membered thiazole ring.</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">four membered cyclic amide fused to a five membered thiazole ring.</span> </div>

answer:
<div class="solution"><p>A penicillin is a member of a family of drugs that have a four membered cyclic amide fused to a five membered thiazole ring. It is narrow spectrum antibiotic.</p><p>Penicillin is an antibiotic or group of antibiotics produced naturally by certain blue moulds, now usually prepared synthetically. </p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/7b96c52e-e034-41d9-8765-0cfb3c98f85c-image.png" style="width: 283px; height: 184px;"/></p></div>


tags: ['Mathematics', 'Permutation Combination', 'BITSAT', 'BITSAT 2020']
source_db: MarksBatch1_P1.db
title:
a-person-invites-a-party-of-10-friends-at-dinner-and-place-them-so-that-4-are-on-one-round-table-and-6-on-the-other-round-table-the-number-of-ways-in-

slug:
a-person-invites-a-party-of-10-friends-at-dinner-and-place-them-so-that-4-are-on-one-round-table-and-6-on-the-other-round-table-the-number-of-ways-in-28034

question:
<div class="question">A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{(10) !}{6 !}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\frac{(10) !}{24}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{(9) !}{24}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">None of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{(10) !}{24}$</span> </div>

answer:
<div class="solution">Selection of 6 guests $=10 \mathrm{C} 6$ 
<br/> <br/>Permutation of 6 on round table $=5 !$ 
<br/> <br/>Permutation of 4 on round table $=3 !$
<br/> <br/>Then, total number of arrangements $=10 \mathrm{C} 6.5 ! .3 !$
<br/> <br/>$=\frac{(10) !}{6 ! 4 !} \cdot 5 ! .3 !=\frac{(10) !}{24}$</div>


tags: ['Mathematics', 'Application of Derivatives', 'BITSAT', 'BITSAT 2009']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">10</span> </div>
source_db: MarksBatch1_P1.db
title:
a-person-standing-on-the-bank-of-a-river-observes-that-the-angle-subtended-by-a-tree-on-the-opposite-bank-is-60circ-when-he-retreats-20-feet-from-the-

slug:
a-person-standing-on-the-bank-of-a-river-observes-that-the-angle-subtended-by-a-tree-on-the-opposite-bank-is-60circ-when-he-retreats-20-feet-from-the-53162

question:
<div class="question">A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is \(60^{\circ}\). when he retreats 20 feet from the bank, he finds the angle to be \(30^{\circ}\). The breadth of the river in feet is :</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">15</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\(15 \sqrt{3}\)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\(10 \sqrt{3}\)</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">10</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

answer:
<div class="solution">et \(h\) be the height of tree PQ and breadth of river PS be \(x \mathrm{~ft}\). Angle of elevation subtended by a tree is \(60^{\circ}\). Also, when he retreats 20 feet, the angle becomes \(30^{\circ}\).<br/><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/jOllLNiEMLkHgFUugK12xCQ7Ec8Jb8rofS_ArBzIMR8.original.fullsize.png"/><br/><br/>Also, in \(\triangle \mathrm{PQS}\),<br/>\(\begin{aligned}<br/>&amp; \tan 60^{\circ}=\frac{h}{x} \\<br/>&amp; \Rightarrow h=\sqrt{3} x<br/>\end{aligned}\)<br/>and in \(\triangle \mathrm{PQR}\),<br/>\(\begin{aligned}<br/>&amp; \tan 30^{\circ}=\frac{h}{x+20} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+20} \\<br/>&amp; \Rightarrow x+20=\sqrt{3} h \\<br/>&amp; \Rightarrow x+20=3 x \Rightarrow 2 x=20 \Rightarrow x=10<br/>\end{aligned}\)<br/>Hence breadth of river is \(10 \mathrm{ft}\).</div>


tags: ['Physics', 'Laws of Motion', 'BITSAT', 'BITSAT 2014']
source_db: MarksBatch1_P1.db
title:
a-person-with-his-hand-in-his-pocket-is-skating-on-ice-at-the-rate-of-10-m-s-and-describes-a-circle-of-radius-50-m-what-is-his-inclination-to-vertical

slug:
a-person-with-his-hand-in-his-pocket-is-skating-on-ice-at-the-rate-of-10-m-s-and-describes-a-circle-of-radius-50-m-what-is-his-inclination-to-vertical-30934

question:
<div class="question">A person with his hand in his pocket is skating on ice at the rate of $10 \mathrm{~m} / \mathrm{s}$ and describes a circle of radius $50 \mathrm{~m}$. What is his inclination to vertical:
<br/> <br/>$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\tan ^{-1}(1 / 2)$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\tan ^{-1}(1 / 5)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\tan ^{-1}(3 / 5)$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\tan ^{-1}(1 / 10)$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\tan ^{-1}(1 / 5)$</span> </div>

answer:
<div class="solution">Since surface (ice) is frictionless, so the centripetal force required for skating will be provided by inclination of boy with the vertical and that angle is given as $\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}$ where $\mathrm{v}$ is speed of skating $\&amp;$
<br/> <br/>i is radius of circle in which he moves.</div>


tags: ['Mathematics', 'Permutation Combination', 'BITSAT', 'BITSAT 2019']
source_db: MarksBatch1_P1.db
title:
a-person-writes-letter-to-six-friends-and-addresses-the-corresponding-envelopes-let-x-be-the-numbers-of-ways-so-that-at-least-two-of-the-letters-are-i

slug:
a-person-writes-letter-to-six-friends-and-addresses-the-corresponding-envelopes-let-x-be-the-numbers-of-ways-so-that-at-least-two-of-the-letters-are-i-74677

question:
<div class="question">A person writes letter to six friends and addresses the corresponding envelopes. Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi></math> be the numbers of ways so that at least two of the letters are in wrong envelopes and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>y</mi></math> be the numbers of ways so that all the letters are in wrong envelopes. Then <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>-</mo><mi>y</mi><mo>=</mo></math></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>719</mn></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>265</mn></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>454</mn></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>720</mn></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>454</mn></math></span> </div>

answer:
<div class="solution">Since only one letters can not be in wrong envelope so that at least two letters are in wrong envelope means all the letters are not in right envelopes.<p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mi>x</mi><mo>=</mo><mn>6</mn><mo>!</mo><mo>-</mo><mn>1</mn><mo>=</mo><mn>720</mn><mo>-</mo><mn>1</mn><mo>=</mo><mn>719</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Y</mi><mo>=</mo></math> number of ways, so that all the letters are in wrong envelops</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>6</mn><mo>!</mo><mfenced><mrow><mn>1</mn><mo>-</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>!</mo></mrow></mfrac><mo>+</mo><mfrac><mn>1</mn><mrow><mn>2</mn><mo>!</mo></mrow></mfrac><mo>-</mo><mfrac><mn>1</mn><mrow><mn>3</mn><mo>!</mo></mrow></mfrac><mo>+</mo><mfrac><mn>1</mn><mrow><mn>4</mn><mo>!</mo></mrow></mfrac><mo>-</mo><mfrac><mn>1</mn><mrow><mn>5</mn><mo>!</mo></mrow></mfrac><mo>+</mo><mfrac><mn>1</mn><mrow><mn>6</mn><mo>!</mo></mrow></mfrac></mrow></mfenced></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>360</mn><mo>-</mo><mn>120</mn><mo>+</mo><mn>30</mn><mo>-</mo><mn>6</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>265</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mi>x</mi><mo>-</mo><mi>y</mi><mo>=</mo><mn>719</mn><mo>-</mo><mn>265</mn><mo>=</mo><mn>454</mn></math></p></div>


tags: ['Physics', 'Dual Nature of Matter', 'BITSAT', 'BITSAT 2021']
source_db: MarksBatch1_P1.db
title:
a-plane-electromagnetic-wave-is-incident-on-a-plane-surface-of-area-a-normally-and-is-perfectly-reflected-if-energy-e-strikes-the-surface-in-time-t-th

slug:
a-plane-electromagnetic-wave-is-incident-on-a-plane-surface-of-area-a-normally-and-is-perfectly-reflected-if-energy-e-strikes-the-surface-in-time-t-th-55631

question:
<div class="question">A plane electromagnetic wave is incident on a plane surface of area A, normally and is perfectly reflected. If energy E strikes the surface in time $t$ then force exerted on the surface is ( $c=$ speed of light)</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\frac{2 \mathrm{E}}{\mathrm{Atc}}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{\mathrm{E}}{2 \mathrm{c}}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\frac{2 \mathrm{E}}{\mathrm{ct}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">zero</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{2 \mathrm{E}}{\mathrm{ct}}$</span> </div>

answer:
<div class="solution">Incident momentum, $p=\frac{E}{c}$
<br/> <br/>For perfectly reflecting surface with normal incidence
<br/> <br/>$\Delta p=2 p=\frac{2 \mathrm{E}}{c} ; \mathrm{F}=\frac{\Delta p}{\Delta t}=\frac{2 \mathrm{E}}{c t}$</div>


tags: ['Physics', 'Dual Nature of Matter', 'BITSAT', 'BITSAT 2020']
source_db: MarksBatch1_P1.db
title:
a-plane-electromagnetic-wave-is-incident-on-a-plane-surface-of-area-a-normally-and-is-perfectly-reflected-if-energy-e-strikes-the-surface-in-timet-the

slug:
a-plane-electromagnetic-wave-is-incident-on-a-plane-surface-of-area-a-normally-and-is-perfectly-reflected-if-energy-e-strikes-the-surface-in-timet-the-72336

question:
<div class="question">A plane electromagnetic wave is incident on a plane surface of area A, normally and is perfectly reflected. If energy E strikes the surface in timet then average pressure exerted on the surface is $(\mathrm{c}=$ speed of light $)$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">zero</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{E} / \mathrm{Atc}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$2 \mathrm{E} / \mathrm{Atc}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{E} / \mathrm{c}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$2 \mathrm{E} / \mathrm{Atc}$</span> </div>

answer:
<div class="solution">Incident momentum, $\mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}}$ For perfectly reflecting surface with normal incidence 
<br/> <br/>$\Delta \mathrm{p}=2 \mathrm{p}=\frac{2 \mathrm{E}}{\mathrm{c}}$
<br/> <br/>$\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{2 \mathrm{E}}{\mathrm{ct}}$
<br/> <br/>$\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{2 \mathrm{E}}{\mathrm{ctA}}$</div>


tags: ['Physics', 'Ray Optics', 'BITSAT', 'BITSAT 2015']
source_db: MarksBatch1_P1.db
title:
a-planoconvex-lens-fits-exactly-into-a-planoconcave-lens-their-plane-surface-are-parallel-to-each-other-if-the-lenses-are-made-of-different-materials-

slug:
a-planoconvex-lens-fits-exactly-into-a-planoconcave-lens-their-plane-surface-are-parallel-to-each-other-if-the-lenses-are-made-of-different-materials-18168

question:
<div class="question">A planoconvex lens fits exactly into a planoconcave lens. Their plane surface are parallel to each other. If the lenses are made of different materials of refractive indices $\mu_{1} \&amp; \mu_{2}$ and $R$ is the radius of curvature of the curved surface of the lenses, then focal length of combination is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\frac{R}{\mu_{1}-\mu_{2}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{2 R}{\mu_{1}-\mu_{2}}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\frac{R}{2\left(\mu_{1}-\mu_{2}\right)}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\frac{R}{2-\left(\mu_{1}+\mu_{2}\right)}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\frac{R}{\mu_{1}-\mu_{2}}$</span> </div>

answer:
<div class="solution">If F be the equivalent focal length, then
<br/> <br/>$$
<br/> <br/>\begin{aligned}
<br/> <br/>\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \quad \frac{1}{F}=\left(\mu_{1}-1\right)\left(\frac{1}{\infty}+\frac{1}{R}\right) \\
<br/> <br/>+\left(\mu_{2}-1\right)\left(\frac{1}{-R}-\frac{1}{\infty}\right)=\frac{\mu_{1}-\mu_{2}}{R} \\
<br/> <br/>F=\frac{R}{\mu_{1}-\mu_{2}}
<br/> <br/>\end{aligned}
<br/> <br/>$$</div>


tags: ['Physics', 'Ray Optics', 'BITSAT', 'BITSAT 2017']
source_db: MarksBatch1_P1.db
title:
a-planoconvex-lens-fits-exactly-into-a-planoconcave-lens-their-plane-surfaces-are-parallel-to-each-other-if-lenses-are-made-of-different-materials-of-

slug:
a-planoconvex-lens-fits-exactly-into-a-planoconcave-lens-their-plane-surfaces-are-parallel-to-each-other-if-lenses-are-made-of-different-materials-of-11892

question:
<div class="question">A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>μ</mi><mn>1</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>μ</mi><mn>2</mn></msub><mo>,</mo><mi>R</mi></math> is the radius of curvature of the curved surface of the lenses, then the focal length of combination is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>R</mi><mrow><mn>2</mn><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>+</mo><msub><mi>μ</mi><mn>2</mn></msub></mrow></mfenced></mrow></mfrac></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>R</mi><mrow><mn>2</mn><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><msub><mi>μ</mi><mn>2</mn></msub></mrow></mfenced></mrow></mfrac></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>R</mi><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><msub><mi>μ</mi><mn>2</mn></msub></mrow></mfenced></mfrac></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>2</mn><mi>R</mi></mrow><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><msub><mi>μ</mi><mn>2</mn></msub></mrow></mfrac></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>R</mi><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><msub><mi>μ</mi><mn>2</mn></msub></mrow></mfenced></mfrac></math></span> </div>

answer:
<div class="solution"><p>Focal length of combination</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mi>f</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><msub><mi>f</mi><mn>1</mn></msub></mfrac><mo>+</mo><mfrac><mn>1</mn><msub><mi>f</mi><mn>2</mn></msub></mfrac><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo>…</mo><mfenced><mi mathvariant="normal">i</mi></mfenced></math></p><p>Using lens maker's formula <br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msub><mi>f</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfenced><mrow><msub><mi>μ</mi><mn>2</mn></msub><mo>-</mo><mn>1</mn></mrow></mfenced><mfenced><mrow><mfrac><mn>1</mn><mrow><mo>-</mo><mi>R</mi></mrow></mfrac><mo>-</mo><mfrac><mn>1</mn><mo>∞</mo></mfrac></mrow></mfenced><mo>=</mo><mo>-</mo><mfrac><mfenced><mrow><msub><mi>μ</mi><mn>2</mn></msub><mo>-</mo><mn>1</mn></mrow></mfenced><mi>R</mi></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msub><mi>f</mi><mn>1</mn></msub></mfrac><mo>=</mo><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><mn>1</mn></mrow></mfenced><mfenced><mrow><mfrac><mn>1</mn><mo>∞</mo></mfrac><mo>-</mo><mfrac><mn>1</mn><mrow><mo>-</mo><mi>R</mi></mrow></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><mn>1</mn></mrow><mi>R</mi></mfrac></math></p><p>or <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msub><mi>f</mi><mn>1</mn></msub></mfrac><mo>=</mo><mfrac><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><mn>1</mn></mrow></mfenced><mi>R</mi></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><msub><mi>f</mi><mn>2</mn></msub></mfrac><mo>=</mo><mo>-</mo><mfrac><mfenced><mrow><msub><mi>μ</mi><mn>2</mn></msub><mo>-</mo><mn>1</mn></mrow></mfenced><mi>R</mi></mfrac></math></p><p><img src="https://cdn-question-pool.getmarks.app/pyq/bitsat/f477415a-ba04-43a0-b7a5-9bdb81b2bf33-image.png" style="width: 180px; height: 242px;"/></p><p>Putting these values in Equation <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mi mathvariant="normal">i</mi><mo>)</mo></math>, we get</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><mn>1</mn><mi>f</mi></mfrac><mo>=</mo><mfrac><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><mn>1</mn></mrow></mfenced><mi>R</mi></mfrac><mo>-</mo><mfrac><mfenced><mrow><msub><mi>μ</mi><mn>2</mn></msub><mo>-</mo><mn>1</mn></mrow></mfenced><mi>R</mi></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><mn>1</mn><mi>f</mi></mfrac><mo>=</mo><mfrac><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><mn>1</mn><mo>-</mo><msub><mi>μ</mi><mn>2</mn></msub><mo>+</mo><mn>1</mn></mrow></mfenced><mi>R</mi></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><mfrac><mn>1</mn><mi>f</mi></mfrac><mo>=</mo><mfrac><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><msub><mi>μ</mi><mn>2</mn></msub></mrow><mi>R</mi></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>f</mi><mo>=</mo><mfrac><mi>R</mi><mfenced><mrow><msub><mi>μ</mi><mn>1</mn></msub><mo>-</mo><msub><mi>μ</mi><mn>2</mn></msub></mrow></mfenced></mfrac></math></p></div>


